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LI  BRARY 

OF   THB 

UNIVERSITY  OF  CALIFORNIA. 


Received  (^pi^^^   ,  fS^^. 

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AN 


ELBMENTAKY     TEEATISE 


ALGEBRA, 


FOR    THE    USE    OF    STUDENTS 


HIGH    SCHOOLS    AND    COLLEGES. 


BY    THOMAS    SHERWIN,    A.   M., 

Principal  of  the  English  High  School,  Boston. 


^  0?  THB 

SIVERSIT??^^^'^^    EDITION. 

t  >5^*/rO».»^    BOSTON. 

HALL    AND    A^^HITING-, 

32  BEOMFIEU)  STREET. 


^.< 


ISntered  according  tc  act  of  Congress,  in  the  year  1841, 

Bv  Thomas  Sherwik, 

In  the  Ck-rk's  Office  of  the  D' strict  Court  of  MaaMcnusetts. 


PREFACE. 


The  author  of  this  treatise  has  endeavored  I:  prepare  a 
work  which  should  sufficiently  exercise  the  ability  of 
most  [earners,  without  becoming,  at  the  same  time,  repul- 
sive to  them  by  being  excessively  abstract.  Some  writers 
err  in  expecting  too  much,  and  others  err,  in  an  equal 
degree,  by  requiring  too  little  of  the  student.  What 
success  has  attended  an  attempt  to  attain  a  proper  medium 
it  is  left  for  competent  teachers  to  decide. 

This  work  commences  in  the  inductive  manner,  be- 
cause that  mode  is  most  attractive  to  beginners.  As  the 
learner  advances,  and  acquires  strength  to  grapple  with  it, 
he  meets  with  the  more  rigorous  kind  of  demonstration. 
This  course  seems  the  most  natural  and  effective.  In- 
duction is  excellent  in  its  place  ;  but  when  an  attempt  is 
made  to  carry  it  into  all  the  departments  of  an  exact 
science,  the  result  often  shows,  that  the  main  object  of 
study  was  misapprehended.  The  young  frequently  fail 
to  deduce  clearly  the  general  principle  from  the  particular 
instances  which  have  engsiged  their  attention. 

Several  parts  of  algebra,  which  are  either  omitted  or 
not  explained  with  sufficient  distinctness  in  other  works, 
liave  received  particular  attention  in  this.  These  partb 
treat  of  principles  and  operations,  with  which  students 
rarely  become  familiar,  but  which  are  essential  to  a  clear 
conjprthension  of  the  subject.  Among  these  operations 
may  be  mentioned  the  separation  of  quantities  into  factors, 
finding  the  divisors  of  quantities,  and  the  substitution  of 
numbers  in  algebraic  fo  mulae. 


ly  PREFACE 

Most  of  the  problems  are  original ;  others  have  teen 
Belected,  which  seemed  the  most  appropriate. 

Although  this  treatise  is  designed  for  students  in  the 
higher  grade  of  seminaries,  it  is  not  beyond  the  reach  of 
any,  who  have  a  good  knowledge  of  arithmetic,  and  who 
are  under  the  guidance  of  Competent  instructors.  Should 
Articles  46,  58,  59,  153  and  154  be  found  too  difficult  for 
the  beginner,  on  his  first  perusal  of  the  book,  they  may 
be  postponed  for  investigation  in  a  review. 

The  writer  is  unwilling  to  close  his  remarks,  without 
expressing  his  obligations  to  others,  who  have  done  so 
much  to  introduce  into  our  country  a  natural  and  rational 
mode  of  studying  mathematics.  Among  these  none  merits 
greater  praise  than  Colburn ;  and  his  works  have  served 
as  a  guide  in  the  composition  of  several  others  on  the 
inductive  plan.  Day,  Smyth,  Davies  and  Peirce  deserve 
also  to  be  mentioned  with  great  respect. 

THOMAS    SHERWIN. 
English  High  School, 
Boston,  Sept.  10,  1841. 


In  this  new  edition  of  his  work,  the  author  would 
remark,  that  the  few  errors  of  the  first  edition  have  been 
carefully  corrected ;  that  a  Key  to  the  Algebra  has  been 
published  ;  and  that,  in  both  the  Algebra  and  Key,  a 
marked  distinction  has  been  made  between  the  full  point 
when  used  as  the  sign  of  multiplication  and  whe*i  used 
EU5  a  decimal  point;  in  the  latter  case,  the  type  being 
inverted,  and  the  sign  consequently  elevated. 

T.   8. 

ApRii  4,  1843. 


CON!  ENTS. 


Preliminary  remarks, 1 

Algebraic  signs, lb. 

Axioms, .' 2 

1      Equations  of  the  first  degree,  having  only  unknown  t^rms  in 

one  member,  and  known  terms  in  the  other, 3 

Definition  of  equation,  member,  term,  coefficient, 4 

II.     Equations  of  the  first  degree  having  unknown  terms  in  one 

member  only,  and  known  terms  in  both, 7 

Transposition  defined  —  rule  tor  transposition, ti — 10 

III.  Equations  of  the  first  degree,  in  which  known  and  unknown 

terms  may  occur  in  each  member, 13 

IV.  Equations  of  the  first  degree,  containing  fractional  parts  of  sin- 

gle terms, 17 

V       Equations  of  the  first  degree,  containing  fractional  parts  of 

quantities  consisting  of  several  terins, 21 

VI.     Equations  of  the  first  degree,  which  require  the  subtraction  of 

quantities  containing  negative  terms, 25 

VL      Multiplication  of  monomials, 29 

Exponents  and  powers  defined, 31 

1^11 1 .     Reduction  of  similar  terms, 33 

IX.     Addition, 35 

X      Subtraction, 37 

Subtraction  of  polynomials  represented, 39 

X I      Multiplication  of  polynomials, 40 

Multiplication  of  polynomials  indicated, 44 

Product  of  sum  and  difierence  —  second  power  of  a  binomial,    45 

Third  power  of  a  binomial, 46 

&  1 1      Division  of  monomials, 47 

Value  of  aO, , 4g 


CONTENTS. 


•V5CT  PAOe 

'   Xlll      Division  of  polynomials, ,.  49 

Infinite  series, 55 

•       Divisibility  ofa'"  — 6"*  by  a  — A, 56 

Conditions  under  which  z*  —  y*  and  z*  +  t/'*  are  divisi- 
ble by  2  -f-  y, 'tS 

Division  of  a  product^by  dividing  one  of  its  factors, b. 

XIV.     Multiplication  of  fractions  by  integral  quantities, o9 

XV.     Division  of  fractions  by  integral  quantities, 64 

XVI.     Factors  and  divisors  of  quantities  —  prime  quantities, 67 

Separation  of  quantities  into  their  factors, 6d 

Finding  ail  the  divisors  of  quantities, 59 

XVII.     Greatest  common  divisor, 71 

X  VllJ      Least  common  multiple, 74 

XIX.     Reduction  of  fractions  to  their  lowest  terms, ...  75 

XX.     Multiplication  of  fractious  by  fractions, 77 

XXI.     Addition  and  subtraction  of  fractions  —  common  denom- 
inator,   79 

XXll.     Division  of  integral  and  fractional  quantities  by  fractions,.  83 

XX  ill      Literal  equations, •      b6 

XXIV.     Equations  of  the  first  degree  with  two  unknown  quantities,  87 

First  method  of  elimination, 90 

Second  method  of  elimination, 92 

Third  method  of  elimination, 93 

XXV.     Equations  of  the  first  degree  with  several  unknown  quan- 
tities,   yfi 

XXVI.     Numerical  substitution  of  algebraic  quantities, 105 

XXVII.     (Generalization, 107 

XXVIU.     Negative  quantities  and  the  interpretation  of  negative  re- 
sults,   119 

XXIX.     Discussion  of  problems, 126 

Signification  of  the  symbol  ^, 129 

Signification  of  the  symbol  Q., iUsJ 

XXX       Extraction  of  the  second  roots  of  numbers, 133 

Irrational  and  imaginary  quantities  defined, 140 

XXXI.     Second  roots  of  fractions  —  and  the  extraction  of  second 

roots  by  approximation,.. i4J 

XXXIl.     Questions  producing  pure  equations  of  the  second  degree,  144 

XXX  111.     Aftected  equations  of  the  s(?cond  degree, \.4f 

XXX  iV.     Extraction  of  the  third  roots  of  numbers, 155 


CONTENTS 


Vll 


ISCT.  PAGB 

XXXV.     Third  roots  of  fractions  —  and  the  extraction  of  third 

roots  by  approximation, 163 

XX XVI.     Questions  producing  pure  equations  of  the  third  degree,  lb6 

XXXVII.     Powers  of  monomials, }G8 

XXXVIIl.     Powers  of  polynomials, 169 

XXXIX.     Binomial  theorem, IT'-i 

XL.     Roots  of  numbers  to  any  degree, 180 

XLI.     Roots  of  monomials, ,  18i\ 

Origin  and  signification  of  fractional  «=  jponents, ISo 

Separation  of  a  quantity  into  any  number  of  factors,  ,  1S6 

XLIl.     Roots  of  polynomials, ib 

XLIII.     Simplification  of  irrational  or  radical  quantities, 191 

XLIV.    Operations  on  irrational  quantities  with  fractional  ex- 
ponents,    ID4 

XLV.     Operations  on  irrational  quantities  with  radical  signs,...  201 

XLVl.     Negative  exponents, 209 

XLVII.    Inequalities, 213 

XLVIII.     Equidifierence, 217 

XLIX.    Ratio  and  proportion, 219 

L.     Progression  by  difference, 229 

LI.     Examples  involving  progression  by  difference, 234 

Lll.     Progression  by  quotient, 236 

LIU.     Examples  in  progression  by  quotient, o . . .  242 

LI  V.     Exercises  in  equations  of  the  second  degree, 244 

LV.     Exercises  in  equations  of  the  second  degree  with  two 

unknown  quantities, 250 

LVI.    Logarithms, 258 

LVII.     Use  of  the  tables  in  finding  the  logarithms  of  given  num- 
bers, and  the  reverse, 265 

l»VIII.     Application  of  logauthms  to  arithmetical  operations,....  278 

LIX     Compound  mtf rest, 282 

LX     Annuities, 288 

Misceilaneou-)  qiiestions 299 


OP  ifffB        >^ 

'UHI7BKSIT71 


ELEMENTS    OF   ALGEBRA. 


PRELIMINARY    REMARKS. 

Art.  1.  Arithmetic  treats  of  numbers  which  have  known  and 
definite  values ;  but  Algebra  makes  use  of  symbols,  which  may 
represent  known,  unknown,  or  indeterminate  quantities.  These 
symbols  are  the  letters  of  the  alphabet. 

Moreover,  in  Arithmetic,  after  an  answer  to  a  question  has 
been  obtained,  it  contains  nothing  in  itself  to  show  by  what 
operations  it  was  found.  For  instance,  suppose  the  number  6  is 
ascertained  to  be  the  answer  to  a  particular  question ;  this  ex- 
hibits no  marks  to  show  whether  it  was  obtained  by  addition, 
multiplication,  division,  or  by  some  other  process  or  combina- 
tion of  processes ;  but  the  results  of  pure  A  Igebra,  that  is,  when 
both  known  and  unknown  quantities  are  represented  by  letters, 
always  irndicate,  or  may  be  made  to  indicate,  the  means  by  which 
Miey  were  produced. 

Algebra  enables  us  also  to  carry  on  a  course  of  reasoning  with 
much  greater  ease  and  rapidity  than  Arithmetic,  and  to  arrive  at 
the  solution  of  problems,  which,  by  the  aid  of  Arithmetic  alone, 
would  be  exceedingly  difficult,  if  not  impossible. 

Art.  ^.  We  proceed  to  notice  some  of  the  signs,  which  mos» 
frequently  occur  in  Algebra. 

The  sign  -\-  is  used  to  express  addition^  and  is  called  plus^ 
which  signifies  more ;  thus,  6  -|-  3  is  read  6  plus  3,  and  means 
I 


PRELIMINARY    RKMARKS 


that  6  and  3  are  to  be  added  together,  or  indicates  the  sum  of  t 
and  3. 

The  sign  —  expresses  subtraction  ^  and  is  called  minus ^  uhicb 
signifies  less ;  thus  8  —  3  is  read  8  minus  3,  and  means  that  3  i? 
to  be  subtracted  from  8,  or  indicates  the  difference  between  8  and  3. 

jMoreover,  quantities  having  before  them  the  sign  -|-,  expressed 
or  understood,  are  called  positive ;  those  having  before  them  iIip 
sign  —    a»-f»  railed  negative  quantitirs. 

Multiplication  is  represented  by  the  sign  X  ;  thus,  6X4 
means  that  6  and  4  are  to  be  multiplied  together,  or  indicates 
the  product  of  6  and  4.  Sometimes  a  point  between  the  quanti- 
fies, as  6  .  4,  has  the  same  signification. 

Division  is  represented  by  the  sign  -i-,  or  :  ;  but  more  fre- 
<^uently  it  is  expressed  in  the  form  of  a  fraction  ;  thus,  6  ~  3,  6  :  3 
and  f ,  each  signifies  the  division  of  6  by  3,  or  indicates  the  quo- 
iient  of  that  division. 

To  express  equality  we  use  two  horizontal  parallel  lines,  thus 
=;z ;  this  is  read  equal  to,  equals,  or  by  some  words  of  similar 
import;  for  example,  6  +  4=  10  means  that  the  sum  of  6  and  4 
is  equal  to  10,  and  is  read  6  plus  4  equals  10. 

Accordingly,  5  X  ^ -\- "t  ==^  ^2^  —  3  means,  that  if  5  be  multi- 
plied by  4,  and  7  be  added  to  the  product,  the  result  will  be  the 
same  as  if  60  be  divided  by  2,  and  3  be  subtracted  from  the  quo- 
tient. 

The  sign  ^,  or  <^,  is  used  to  express  the  inequality  of  quan- 
tities ;  thus,  8  >  5,  or  5  <^  8,  signifies  that  8  is  greater  than  5, 
or  that  5  is  less  than  8,  the  open  end  always  being  placed 
towards  the  greater  quantity. 

To  represent  unknown  quantities,  we  use  some  of  the  last  let- 
ters of  the  alphabet,  as  x,  y,  &c. ;  and  to  represent  known  quan 
titles,  we  use  some  of  the  first  letters,  as  a,  b,  c,  &c. ;  althouj^h, 
in  many  problems  of  this  work,  known  quantities  are  expressed 
by  figures. 

Art.  3.  There  are  some  propositions,  the  truth  of  which  ia 
manifest,  as  soon  as  they  are  presented  to  the  mind.  These  pro- 
liositions  are  called  axioms ;  the  following  are  of  this  kind. 


AXIOMS.       EQUATIONS    OF    THE    FIRST    DKGBEF. 


1.  If  the  same  quantity  or  equal  quantities  be  added  to  equa. 
quantities,  the  sums  will  be  equal. 

2.  If  the  same  quantity  or  equal  quantities  be  subtracted  from 
equal  quantities,  the  remainders  will  be  equal. 

3.  If  equal  quantities  be  midtiplicd  by  the  same  quantity  or  by 
equa.  qiantities,  the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same  quantity  or  by 
equal  quantities,  the  quotients  will  be  equal. 

5.  If  the  same  quantity  be  both  added  to  and  subtracted  from 
another,  the  value  of  the  latter  will  not  be  changed. 

6.  If  a  quantity  be  both  multiplied  and  divided  by  another,  ita 
value  will  not  be  changed. 

7.  Two  quantities,  each  of  which  is  equal  to  a  third,  are  equal 
to  each  other. 

8.  The  whole  of  a  quantity  is  greater  than  a  part  of  it 

9.  The  whole  of  a  quantity  is  equal  to  the  sum  of  all  its  par^s 


SECTION    I. 

EQTTATIOIfS    OF    THE  FIRST  DEGREE,  HAVING  OJVLY  UNKNOWN  TERMS 
IN    ONE    MEMBER    AND    KNOWN    QUANTITIES    IN    THE    OTHER. 

Art.  4:.  1.  An  apple  and  an  orange  together  cost  6  cents;  but 
the  orange  cost  twice  as  much  as  the  apple.  What  was  the  price 
of  each  1 

In  this  question,  if  we  knew  the  price  of  the  apple,  we  should, 
by  doubling  it,  obtain  that  of  the  orange.  The  price  of  the 
apple,  then,  may  be  considered  as  the  unknown  quantity. 

Suppose  that  X  represents  the  number  of  cents  given  for  the 
apple ;    twice  as  much,  or  the  price  of  the  orange,  would   je 
represented  by  2  x. 
Hence,     z  -j-  2  x  =  6.     Putting  the  z's  together,  we  have 
8  z  =  6 ;  one  x  will  be  ^  as  much: 
therefore,      x  =  2  cents  =  tne  price  ol  ihe  apple ; 
and  2  x  =z  4  cents  =  the  price  of  the  orange. 


4        EQUATIONS    OF    THE    FIRST    DEGREE.        DEFINITIONS,    ETC.        L 

Remark.  Questions  in  Algebra  may  be  proved  as  uell  as 
ihose  in  Arithmetic.  The  proof  of  the  foregoing,  would  consist 
in  adding  the  price  of  the  apple  to  that  of  the  orange,  and  ascer- 
taining that  their  sum  is  6  cents.  Let  the  learner  prove  the  cor- 
rectness of  his  answers,  as  he  advances. 

A  representation  of  the  equality  of  quantities,  is  called  an 
t'juation.     Thus,  x-\-2x=zfy  is  an  equation. 

A  member  or  side  of  an  equation,  signifies  the  quantity  or 
quantities  on  the  same  side  of  the  sign  =,  theirs/  member  being 
<5ii  the  left,  and  the  second  member  on  the  right  hand  side  of  this 
sign. 

An  equation  of  the  first  degree  is  one,  in  which  the  unl  nown 
quantities  are  neither  multiplied  by  themselves  nor  by  each 
other. 

The  separate  parts  of  an  algebraic  expression  affected  by  the 
signs  -\-  and  — ,  are  called  terms.  Those  terms  which  have  no 
sign  prefixed  to  them,  are  supposed  to  have  the  sign  -|-»  ^nd  a 
quantity  is  said  to  be  affected  by  a  sign,  when  it  is  immediately 
preceded  by  that  sign,  either  expressed  or  understood.  When 
the  first  term  of  a  member  of  an  equation,  or  of  any  algebraic 
quantity,  is  affected  by  the  sign  +,  it  is  usual  to  omit  writing  the 
sign  before  that  term;  but  the  sign  —  must  always  be  written 
before  any  term  affected  by  it.  The  equation,  x  -)-  2  a;  n:  (^^ 
consists  of  three  terms,  two  in  the  first  member  and  one  in  -ne 
second,  and  each  of  these  terms  is  affected  by  the  sign  -\-. 

The  number  written  immediately  before  a  letter,  showing  how 
many  times  the  letter  is  taken,  is  called  the  coefficient  of  that 
letter;  thus,  in  the  expressions,  ^z,  5a:,  7a;,  the  coefficients  of  x 
are  -J,  5  and  7.  A  letter  which  has  no  number  written  before  it, 
is  supposed  to  have  1  for  its  coefficient ;  thus,  x  is  the  same  as  1  a. 
Letters,  as  we  shall  see  hereafter,  may  be  used  as  coefficients. 

'J'he  i>,  ocess  by  which  an  equation  is  formed  from  the  condi 
lions  of  a  (question,  is  called  putting  the  question  into  an  equa- 
Hon  ;  and  the  process  by  which  the  value  of  the  unknown  quan- 
tity is  found  from  the  equation,  is  called  solving  the  equation. 
2    Said  A  to  B,  my  horse  and  saddle  are  worth  $110 ;  but  mj 


I  EQUATIONS    OF    THE    FIRST    UEf^REE.  5 

horse  is  worth  10  times  as  much  as  my  saddle      Required  the 
(vorth  of  each. 

3.  A  man  bought  some  corn  and  rye  for  60  shillings,  the  corn 
at  4s.  per  bushel  and  the  rye  at  6s.,  and  there  was  the  same 
number  of  bushels  of  each.  How  many  bushels  were  there  of 
each? 

Let  z  represent  the  number  of  bushels  of  each  ;  then  x  hushelg 
of  corn  at  4s.  per  bushel,  will  come  to  4  x  shillings,  and  %  bush- 
els of  XY^  at  6s.  per  bushel,  will  come  to  6  x  shillings.  Hence, 
4  2  +  6  X  =  60. 

4.  A  man  sold  an  equal  number  of  oxen,  cows  and  sheep;  the 
oxen  at  $40  apiece,  the  cows  at  $15,  and  the  sheep  at  $5;  the 
whole  came  to  $660.     How  many  were  there  of  each? 

5.  A  woman  bought  some  peaches,  pears  and  melons  for 
$110;  the  peaches  at  I  cent  apiece,  the  pears  at  2,  and  the 
melons  at  12 ;  there  were  twice  as  many  pears  as  melons,  and 
three  times  as  many  peaches  as  pears.  How  many  were  there  of 
each  ? 

Let  X  represent  the  number  of  melons ;  then  2  x  will  represent 
the  number  of  pears,  and  6x,  the  number  of  peaches.  At  1  cent 
each,  6x  peaches  come  to  6x  cents,  2x  pears  at  2  cents  each 
will  come  to  4  x  cents,  and  x  melons  at  12  cents  each  will  come 
to  1 2  /  cents ;  hence,  6  x  -(-  4  x  -|-  12  x  z=  110. 

6.  A  gentleman  hired  a  man  and  a  boy  to  work  a  certain 
number  of  days,  the  man  at  8s.  and  the  boy  at  4s.  per  day,  and 
paid  them  $30.  How  many  days  were  they  employed,  and  how 
much  did  each  receive  ? 

7.  Three  numbers  are  in  the  proportion  of  1,  2  and  3,  and  the 
Bum  of  them  is  630.     What  are  these  numbers  ? 

The  proportion  of  1 ,  2  and  3,  means  that  the  second  is  twice 
and  the  third  three  times  as  much  as  the  first. 

8.  Divide  100  into  three  parts,  in  the  proportion  of  5,  7  and  8. 
The  proportion  of  5,  7  and  8,  means  that  the  2d  is  -^,  and  the 

'5d  I  as  much  as  the  1st. 

Suppose  tne  1st  part  z=:  5x,  then  the  2d  will  be  7x,  an^  the 
W,  8x 

I* 


6  EQUATIONS    OF    THE    FIRST    DEGREE  1 

9.  Two  persons  set  out  at  the  same  time  from  two  towns  150 
miles  apart,  and  travel  towards  each  other  till  they  meet,  one  at 
8  miles  an  hour,  and  the  other  at  7.  How  many  hours  will  they 
be  on  the  road,  and  how  far  will  each  travel  ? 

JO.  Three  robbers,  having  stolen  48  guineas,  quarrelled  about 
the  division  of  them,  and  each  took  as  nmch  as  he  could  get ; 
the  first  obtained  a  certain  sum,  the  second  twice  as  much,  and 
the  third  as  much  as  both  the  others.  How  many  guineas  did 
each  obtain  ? 

11.  A  gentleman  wished  to  divide  an  estate  of  $81000  be- 
tween his  wife  and  two  sons,  so  that  his  wife  should  have  $4,  as 
often  as  the  elder  sou  had  $3,  and  the  younger  $2.  How  much 
^ould  each  receive? 

12.  A  fortress  has  a  garrison  of  1200  men,  a  certain  portion 
of  whom  are  cavalry,  three  times  as  many  artillerymen,  and  six 
times  as  many  infantry.     How  many  are  there  of  each  corps  ? 

13.  In  fencing  a  field,  three  men.  A,  B  and  C,  were  employed 
A  could  fence  9  rods  a  day,  B  7,  and  C  5 ;  B  wrought  twice  as 
many  days  as  A,  and  C  five  times  as  many  as  B.  The  distance 
round  the  field  was  584  rods.  How  many  days  did  each  work, 
and  how  many  rods  of  fence  did  each  build  ? 

14.  A  man  bought  three  pieces  of  cloth  for  $280.  The  sec- 
ond piece  was  twice  as  long  as  the  first,  and  the  third  was  as  long 
as  the  first  two.  He  gave  $4  a  yard  for  the  first  piece,  $5  a  yard 
for  the  second,  and  $7  a  yard  for  the  third.  Required  the  num- 
ber of  yards  in  each  piece. 

15.  Four  cows,  3  calves  and  10  sheep  cost  $112.  A  cow  cost 
5  times  as  much  as  a  calf,  and  a  calf  cost  twice  as  much  as  a 
sheep.     Required  the  price  of  each. 

16.  A  cistern  holding  140  gallons,  was  filled  with  water  by 
means  of  two  buckets,  the  greater  of  which  held  twice  as  much 
as  the  less.  The  greater  was  emptied  7  times  and  the  less  6 
times.     How  many  gallons  did  each  bucket  holdt 

17.  A  boy  being  sent  to  market,  bought  some  beef  at  14  cents 
A  pound,  and  twice  as  much  mutton  at  9  cents  a  pound.    He  was 


(I  EQUATIONS    OF    THE    FIRST    DEGREE.  1 

intrusted   with    $4    and   brought   back   80  cents.     How    n.ny 
pounds  of  each  kind  of  meat  did  he  buy? 

18.  A  man  wished  to  pay  $60,  with  dollars,  halves,  quarterSj 
and  eighths,  of  each  an  equal  number.  How  many  coins  of  each 
kind  would  he  require? 

19.  A  man  paid  c£l44  in  guineas  at  21s.  and  crowns  at  5s.  each. 
There  were  three  times  as  many  crowns  as  guineas.  Required 
the  number  of  each. 

20.  A  man  on  a  journey  traveled  twice  as  far  the  2d  day  as 
he  did  the  Ist ;  on  the  3d  day,  as  far  as  he  did  the  first  two  days ; 
on  the  4th  day,  as  far  as  he  did  the  first  three  days ;  and  on  the 
5th  day,  half  as  far  as  on  the  4th.  The  whole  distance  traveled 
was  150  miles.     How  far  did  he  go  each  day  ? 

21.  A  merchant  exchanged  rye  at  7s.  and  wheat  at  9s.  a  bushel, 
of  each  the  same  quantity,  for  32  bushels  of  corn  at  4s.  a  bushel. 
How  many  bushels  of  rye  and  wheat  were  given  in  exchange  ? 

22.  A  drover  bartered  6  oxen  and  10  cows  for  a  farm  of  50 
acres  at  $1J  per  acre.  He  reckoned  each  ox  worth  as  much  as 
two  cows.  What  price  was  assigned  to  an  ox  and  a  cow  re- 
spectively ? 

Art.  o.  In  the  preceding  questions,  ar's,  that  is,  unknown  quan- 
tities, have  been  found  only  in  the  first  member  of  the  equation, 
and  they  have  all  been  affected  by  the  sign  -|- ;  and  we  perceive, 
that,  after  an  equation  was  formed,  the  first  step  was  to  reduce 
or  combine  all  the  unknown  quantities  into  one  term,  which  is 
done  by  adding  the  coefficients ;  after  which,  the  value  of  the 
unknown  quantity  was  found  by  dividing  both  members  by  the 
eoefficient  of  the  unknown  quantity. 


SECTION    II. 

EQUATION'S    OF    THE     FIRST    DEGREE,    HAVING    UNKNOWN    TERMS     IM 
ONE    MEMBER    ONL.Y,    AND     KNOWN    TERMS    IN    BOTH    MEMBERS. 

Art.  6.  Two  brothers  had  together  $20,  but  the  elder  had  two 
dollars  more  than  the  younger.     How  much  money  had  each? 


S  EQUATIONS    OF    THE    FIRST    DEGREE.  II 

Let  X  represent  the  number  of  dollars  the  younger  had; 
then  X  -f-  2  =        the  number  of  dollars  the  elder  had ; 
consequently,  z  -|-  z  -f-  2  =:  20 ;  or,  combining  the  z's, 
2x  +  2  =  20. 
Now  as  the  two  members  are  equal,  we  can  subtract  2  from 
each,  and  the  remainders  will  be  equal  (ax.  2,  Art.  3);  2  sub 
tracted  from  2  z  -|-  2,  leaves  2  z,  and  2  subtracted  from  20,  leaves 
18;  hence,     2z=  18; 

zrz:  9,  number  of  dollars  the  younger  had, 
and  z  -|-  2  r=  11,  number  of  dollars  the  elder  had. 
Instead  of  actually  subtracting  2  from  the  second  member  at 
once,  we  may  subtract  it  from  the  first,  and  represent  it  as  sub- 
tracted from  the  second;  thus,  2z  =  20  —  2;  now  performing 
the  subtraction  indicated,  we  have  2z=:  18,  the  same  as  before. 
The  equation  2zzz:20  —  2  is  obtained  from  2z  +  2z=20  merely 
by  removing  the  2  to  the  other  side  of  the  sign  rz:,  and  changing 
its  sign  from  -(-  to  — . 

Art.  7.  Removing  a  term  from  one  member  of  an  equation  to 
the  other,  is  called  transposing  that  term,  or  transposition.  Any 
term^  therefore^  affected  by  the  sign  -(-,  may  be  transposed^  if 
this  sign  be  changed  to  — . 

1.  Two  men,  A  and  B,  hired  a  house  for  $650,  of  which  A 
paid  $150  more  than  B.     What  did  each  pay? 
Let  X  represent  the  number  of  dollars  B  paid. 
Then  z  -|-  150  will  represent  the  number  A  paid. 
Hence,     z -(- z -f- 1 50  rr  650.     Reducing, 

2  z-j- 150  =  650;  transposing  150, 
2  z  =  650  —  150 ;  reducing  the  2d  member, 
2z  =  500, 
X  =  $250  =  what  B  paid, 
z  +  150  z=  $400  =  what  A  paid. 
2    Two  men  possess  together  $56000,    but  the  second   has 
HOOOO  more  than  the  first.     How  much  money  has  each? 

3.  Two  towns  are  at  unequal  distances  and  in  opposite  direc- 
tions from  Boston ;   the  distance  between  th  ese  towns   is   23(' 


U.  EQUATIONS    OF    THE    FIRST    DEGREE.  9 

miles,  but  one  is  10  miles  more  than  twice  as  far  from  Boston  aa 
the  other.     What  is  the  distance  of  each  from  that  city  ? 

4.  The  sum  of  the  ages  of  A,  B  and  C  is  100  years ;  but  B's 
age  is  twice  that  of  A  and  5  years  more,  and  C's  age  is  equal  to 
the  sum  of  A's  and  B's.     Required  the  age  of  each. 

5.  A  grocer  wishes  to  make  a  mixture  of  four  kinds  of  tea,  so 
that  there  shall  be  6  lbs.  more  than  twice  as  much  of  the  2d  kind 
as  of  the  1st,  as  many  lbs.  of  the  3d  as  there  are  of  the  first  two, 
and  as  many  of  the  4th  as  there  are  of  all  the  others ;  the  whole 
mixture  is  to  contain  120  lbs.  How  many  lbs.  must  there  be  of 
fiach  sort? 

6.  A  man  has  five  sons,  each  of  whom  is  two  years  older  than 
his  next  younger  brother,  and  the  amount  of  their  ages  is  50  years 
What  is  the  age  of  each  ? 

7.  A  merchant  bought  10  pieces  of  cloth  for  $331 ;  5  pieces 
were  blue,  3  green,  and  2  black ;  a  piece  of  green  cost  $2  more 
than  one  of  black,  and  a  piece  of  blue  $3  more  than  one  of 
green.     How  much  did  each  kind  cost  per  piece  ? 

8.  Says  A  to  B,  my  age  is  10  years  more  than  yours,  and 
twice  my  age  added  to  three  times  yours,  makes  120  years.  Re- 
quired the  age  of  each. 

9.  A  gentleman  leaves  an  estate  of  $10000,  to  be  divided  be- 
tween his  three  daughters  and  two  sons,  in  the  following  man- 
ner, viz :  the  daughters  are  all  to  share  equally,  but  the  elder 
son  is  to  have  $1000  more  than  twice  as  much  as  the  younger 
and  the  younger  exactly  twice  as  much  as  one  of  the  daughters 
What  is  the  share  of  each  ? 

10.  A  laborer  undertook  to  reap  6  acres  of  wheat  and  10  acres 
of  oats  for  $21f ,  or  130  shillings ;  but  he  was  to  have  3s.  more 
an  acre  for  the  wheat  than  for  the  oats.  What  was  the  price  of 
reaping  an  acre  of  each  ? 

Let  X  shillings  represent  the  price  of  reaping  the  wheat  per  acre 
Then  x  —  3  will  be  the  price  of  reaping  the  oats  per  acre. 
Six  acres  of  wheat  will  cost  (i  x  shillings ; 
and  ten  acres  of  oats  will  cost  10  a:  —  30  shillings. 


10  EQUATIONS    OF    THE    FIRST    DEUKEE.  U 

Hence,  6x+10x  —  30r=  130.     Reducing, 

16x  — 30=:U0. 
By  adding  30  to  each  member  (ax.  1,  Art.  3),  the  equation 
becomes 

16x  — 30  +  30  =  130  +  30,  or, 

16  X  =  130  +  30,  since  16  a:  —  30  +  30  is  the  same  as  16  » 
(ax    5) ;  hence,  16  a;  =  160, 
herefore,  x  =.    10, 

and  X  —  3  =:      7.    Ans.  wheat  10s.,  oats  7s.  per  acre 
Most  of  the  preceding  questions  in  this  section,  may  be  solved 
»n  a  similar  way. 

The  equation  162  =  130  +  30  is  obtained  from  16  x  —  30  = 
130,  merely  by  removing  the  30  to  the  second  member  of  the 
equation,  and  changing  its  sign  from  —  to  -\-. 

Art.  8.  HencCf  any  term  affected  by  the  sign  — ,  may  be  trans 
posed  from  one  member  to  the  other  ^  if  its  sign  be  changed  to  -{- ; 
for,  this  is  adding  the  same  quantity  to  each  member  (ax   1). 
This  principle,  together  with  that  established  in  Art.  T,  gives  the 
following  general 

RULE    FOR    TRANSPOSITION. 

Art.  O.  Any  term  may  be  transposed  from  one  member  oj  an 
equation  to  the  other,  care  being  taken  to  change  its  sign  from  — 
to  -}-)  or  from  -j-  to  — . 

It  may  be  remarked,  that  the  value  of  every  such  expression  as, 
1  —  1,  2  —  2,  3  —  3,  &-C.,  or  x — x,  4x  —  4x,  a  —  a,  5  a — 5«, 
&.C,,  is  0  or  nothing;  that  is,  the  plus  and  minus  quantities  equal 
in  value  cancel  each  other. 

Moreover,  when  quantities  are  connected  by  the  signs  -\-  and 
— ,  it  is  of  no  importance  in  what  order  they  stand,  provided 
ihe}  have  their  proper  signs  prefixed  to  them  ;  thus,  3  -J-  7  —  2 
may  be  written  7  -|-  3  —  2,  or  —  2  -|-  7  +  3,  the  value  of  each 
expression  being  8. 

When  the  first  term  is  affected  by  the  sign  +,  it  is  usual  to 

v/uiit  writing  that  sign  ;  but  the  sign  —  must  never  be  omitted 

The  learner  cannot  be  too  careful  with  regard  to  the  signs^ 

a  mistake  in  the  sign  occasions  an  error  equal  to  twice  the 


11.  EQUATIONS    OF    THE    FIRST    DEGREE.  LI 

value  ol'  the  term  aflfected  by  it ;  thus,  12  +  3  is  equal  to  15,  and 
12  —  3  is  equal  to  9 ;  now  the  diffe  ence  between  15  and  9  is  G 
or  twice  3. 

We  perceive  also,  that  when  a  quantity  consisting  of  severa' 
terms,  as  x  —  t^,  is  to  be  multiplied,  each  term  must  be  multi- 
plied and  the  same  signs  retained;  thus  10  times  x — 3  is  10a; — 
30 ;  in  like  manner,  7  times  12  —  32;  is  84  —  21  x. 

1.  At  a  certain  election,  two  persons  were  voted  for;  but -^ he 
candidate  chosen  had  a  majority  of  87,  and  the  whole  numjer 
of  votes  was  899.     How  many  votes  had  each? 

2.  In  a  manufactory,  205  persons,  men,  boys  and  girls,  are 
employed ;  there  are  four  times  as  many  boys  as  men,  and  20  less 
than  ten  times  as  many  girls  as  boys.  How  many  of  each  are 
employed  1 

3.  A  general,  on  reviewing  his  troops,  found  he  had  in  all  2300 
men,  of  whom  a  certain  portion  were  cavalry,  three  times  as 
many  riflemen,  and  100  less  than  four  times  as  many  infantry  as 
riflemen.     How  many  were  there  of  each? 

4.  Four  men,  A,  B,  C  and  D,  enter  into  partnership.  A  con- 
tributes a  certain  sum,  B  three  times  as  much,  C  twice  as  much 
as  A  and  B  both,  and  D  as  much  as  the  other  three  wanting 
$1000.  The  whole  sum  invested  was  $65000;  how  much  did 
each  put  in  trade  ? 

5.  Divide  $491  among  three  persons.  A,  B  and  C,  so  that  A 
shall  have  $270  more,  and  B  $100  less  than  C. 

Suppose  X  =.  C's  share.  Then  x  -\-  270  =  A's  share,  and 
»  —  100  =  B's  share.     Hence,  x  +  2  +  270  +  a;  —  100  =  491 

Reducing,  we  have  3x-|- 170  =  491 ;  for  adding  270  and 
Bubtracting  100,  is  the  same  as  adding  their  difference. 

6.  A  man  aged  80  years,  had  spent  a  certain  part  of  his  life 
in  France,  three  times  as  much  and  30  years  more  in  England 
and  twice  as  much  wanting  10  years  in  America.  How  mary 
years  had  he  lived  in  each  country  ? 

7.  A  certain  town  contains  2900  inhabitants,  English,  Irish, 
and  French ;  there  are  600  fewer  Irish  than  English,  and  400 
fewer  French  than  Irish.     How  many  are  there  of  each? 


12  EQUATIONS    OF    THE    FIRST    DEGREE.  (I 

Let       X  =z  the  number  of  English ; 

then     X  —  600  =  the  number  of  Irish, 

and      X  —  600  —  400  z=.  the  number  of  French. 

Hence,  a:  +  a;  —  600  +  z  —  600  —  400  =  2900 ;  by  reducing 
we  have  3x — 1600  =  2900;  for  all  three  of  the  numbers  af- 
fected by  the  sign  — ,  are  considered  as  separately  subtracted 
and  it  would  evidently  be  the  same  thing  to  subtract  the  sum  of 
them  at  once. 

8.  In  a  casket  containing  390  coins,  there  is  a  certain  num- 
ber of  eagles,  10  less  than  twice  as  many  half-eagles,  and  20  less 
than  three  times  as  many  dollars  as  there  are  half-eagles.  How 
many  coins  are  there  of  each  kind  ? 

9.  A  merchant  bought  a  certain  number  of  yards  of  broad- 
cloth at  $8  per  yard,  6  less  than  three  times  as  many  yards  of 
cassimere  at  $4  per  yard,  and  twice  as  much  silk  at  $1  per  yard 
as  there  were  yards  of  cassimere.  The  whole  came  to  $1264 
How  many  yards  of  each  kind  did  he  buy? 

10.  A  man,  engaged  in  trade,  gained,  the  first  year,  $500 ; 
the  second  year  he  doubled  what  he  then  had ;  but  the  third  year 
he  lost  $2000,  when  it  appeared  that  he  had  remaining  $3000. 
How  much  money  had  he  at  first? 

Suppose   X  ■=.  his  money  at  first. 

Then        x  +    500  =  his  money  at  the  end  of  the  1st  year ; 

2  x  4-  1000  =  his  money  at  the  end  of  the  2d  year,    • 
and        22;+  1000  — 2000  =  his  money  at  the  end  of  the  3d  year 
Hence,  2  x  +  1000  —2000  =  3000 ;  or  reducing, 

2ar— 1000=^3000;  for  2  r  +  1000  —  2000,  signifies 
that  1000  is  added  to  2  x,  and  from  the  sum  2000  is  subtracted, 
which  is  the  same  thing  as  subtracting  1000. 

11.  An  inheritance  of  $92500  is  to  be  divided  among  fi\G 
heirs,  A,  B,  C,  D  and  E,  in  the  following  manner,  viz :  B  is  to 
have  $600  more  than  A ;  C  twice  as  much  as  B,  wanting  $400 ; 
D  as  much  as  A  and  B  both,  wanting  $300 ;  and  E  $500  more 
•ban  A  and  D  both.     What  is  the  share  of  each  ? 


[II.  EQUATIONS    OP    TtiE    FIRST    DEGREE  I? 

Suppose  X  =  the  share  of  A 

Then      x  +    000  =  the  share  of  B, 

2  a:  4-  1200  —  400  =  the  share  of  C, 

a;  _j_  X  +  600  --  300  =  the  share  of  D, 

and         X  +  a:  +  X  -f  600  —  300  +  500  =  the  share  of  E. 

Hence,  x  +  x +  600  +  2  x-{- 1200  —  400  +  z  +  x  +  600  — 
300  _|.  X  _|_  x  -|-  z  -f.  600  —  300  +  500  =  92500.  Reducing,  9 1 
■\-  2500  zzz  92500 ;  for  the  sain  of  the  numbers  affected  by  the 
sign  +  is  3500,  and  the  sum  of  those  affected  by  the  sign  —  is 
1000;  but  adding  3500  and  subtracting  1000  is  the  same  a» 
ndding  2500. 

Had  the  sum  of  the  numbers,  affected  by  the  sign  — ,  been 
greater  than  that  of  the  numbers,  affected  by  the  sign  +,  the  dif- 
ference of  these  two  sums  would  have  had  the  sign  — . 

In  the  above  question,  the  labor  would  have  been  abridged,  if 
the  expressions  for  the  several  shares  had  been  reduced,  as  far  as 
possible,  previous  to  forming  the  equation. 

12.  A  drover  has  a  certain  number  of  oxen;  three  times  as 
many  cows,  wanting  25;  just  as  many  calves  as  cows;  and  100 
more  sheep  than  he  has  oxen  and  cows  together.  The  number 
of  the  whole  is  905 ;  how  many  of  each  has  he  ? 

13.  In  a  company  of  140  persons,  consisting  of  officers,  mer- 
chants and  students,  there  were  4  times  as  many  merchants  as 
students,  wanting  25 ;  and  5  more  than  3  times  as  many  officers 
as  students.     How  many  were  there  of  each  class  ? 


SECTION    111. 

tQITATIONS    OF    THE     FIRST     DEGREE,     IN    WHICH    BOTH    KNOWN    AND 
UNKNOWN    TERMS    MAY    OCCUR    IN    EACH    MEMBER. 

Art.  lO.  What  number  is  that  to  which  if  18  be  added,  thf 
sum  will  be  equal  to  four  times  the  number  itself? 

Let  x  represent  the  number ;  then  x-|-18z=4x,  or  4ar  =  z-f- 
18 ;  as  it  is  evidently  indifferent  which  quantity  is  made  the  firsl 
member. 

2 


14  EQUATIONS    OF    THE    FIRST    DEGREE.  Ill 

Now  it  IS  our  object  to  make  all  the  x's,  or  unknown  quantif 
des,  stand  in  one  member  of  the  equation,  and  the  known  quan 
tities  in  the  other ;  and,  for  the  sake  of  uniformity,  we  gerjcrally 
collect  the  unknown  quantities  into  the  first  member. 

In  the  equation  4x=:x-\-  18,  by  transposing  the  x  from  tlw 
second  member  to  the  first,  that  is,  by  subtracting  x  from  both 
members,  we  have  4  x  —  x=z  18,  or  reducing, 
3  a:  =18,  and 
xz=i6,  Ans. 

Or  we  might  have  taken  the  equation  z  -(-  18  =  4  x  j 

by  transposing  the  18,  we  have  x  =  4  x  — 18 ;  then, 

by  transposing  the  4  x,  we  have  x  —  4xz=z  —  18 ; 

reducing,  —  3  x  =  — 18. 

Here  both  members  are  wholly  minus,  but  by  transposing  both, 
we  have  18  =  3x,  which  is  the  same  as 

3x:=  18;  hence, 
x  =  6. 

The  equation,  3x=  18,  might  have  been  obtained  from  — 3z 
=  —  18,  merely  by  changing  the  signs  to  -|-. 

In  like  manner,  in  the  equation  3x  —  5x  =  20  —  46,  which 
reduced  gives  —  ^xz=z — 26,  we  might  change  all  the  signs  be- 
fore reducing,  which  would  give  —  3x-|-5x  =  —  20  +  46,  or 
2  X  =  26  and  X  =  13. 

Art.  11.  Hence,  the  signs  of  all  the  terms  in  both  members  oj 
an  equation  may  bt  changed ;  for  this  is  the  same  as  transposing^ 
all  these  terms. 

This  change  of  signs  should  be  made,  whenever  the  first  mem- 
ber becomes  minus ;  but  the  learner  must  recollect,  that  terms 
having  no  sign,  are  supposed  to  have  -f-,  and  that  he  must  change 
all  the  signs,  otherwise  great  errors  will  ensue. 

1.  Says  A  to  B,  if  to  my  age  twice  my  age  and  30  years  more 
be  added,  the  sum  will  be. five  times  my  age.     How  old  is  he? 

Let  X  =  his  age ; 

then  6x  =  x  +  2x  +  30.     Reducing  the  2d  member, 

5  X  =  3  X  +  30 ;  transposing  3  x, 


III.  EQUATIONS    OF    THE    FIRST    DEGREE.  15 

6x  —  3  a;  =  30;  reducing, 
2z=r30,    and 
z  =  15  years.     Ans. 

2.  A  merchant  sells  two  kinds  of  cloth,  the  finer  at  $2  a  yard 
more  than  the  coarser;  12  yards  of  the  coarser  come  t)  as  much 
Bs  8  yards  of  the  finer.     What  is  the  price  of  each  per  yard? 

3.  Says  A  to  B,  four  times  my  age  is  equal  to  five  times  yours, 
and  the  difference  of  our  ages  is  10  years.  What  is  the  age  of* 
ftach  ? 

4.  A  man  having  a  certain  number  of  cows  and  the  same 
number  of  sheep,  bought  4  more  cows  and  16  more  sheep ;  he 
then  found  that  three  times  his  number  of  cows  was  equal  to 
twice  his  number  of  sheep.     How  many  had  he  of  each  at  first  ? 

5.  A  father  distributed  a  certain  sum  of  money  among  his  four 
sons.  The  third  received  9d.  more  than  the  youngest ;  the  sec- 
ond, I2d.  more  than  the  third;  and  the  eldest,  18d.  more  than 
the  second.  The  whole  sum  was  6d.  more  than  seven  times 
what  the  youngest  received.  How  much  had  each,  and  what 
was  the  whole  sum  distributed  ? 

6.  A  sum  of  money  was  to  be  divided  among  six  poor  persons, 
so  that  the  second  should  have  3s.,  the  third  2s.,  the  fourth  5s., 
the  fifth  7s.,  and  the  sixth  8s.,  less  than  the  first.  Now  the  sum 
divided  was  7s.  more  than  four  times  the  share  of  the  second. 
What  did  each  receive  ? 

7.  A  person  bought  two  casks  of  beer,  one  of  which  held 
twice  as  much  as  the  other  •  from  the  larger  he  drew  out  20,  and 
fiom  the  smaller  25  gallons,  he  then  found  that  there  remained 
in  the  larger  4  times  as  much  as  in  the  smaller.  What  did  each 
cask  contain  at  first? 

8.  A  man  bought  10  bushels  of  wheat  and  16  bushels  of  rye, 
the  wheat  cost  2s.  more  per  bushel  than  the  rye,  and  the  whole 
cost  of  the  wheat  wanted  16s,  to  be  equal  to  that  of  the  rye. 
What  was  the  price  of  each  per  bushel  ? 

9.  An  instructor,  wishing  to  arrange  his  pupils  in  rows  with  a 
certain  number  in  each  row,  found  that  there  were  3  too  many 
to  make  six  rows,  and  4  too  few  to  make  seven  rows.     How 


16  EQUATIONS    OF    THE    FIRST    DEGREE.  HI 

many  did  he  wish  to  place  in  a  row,  and  how  many  scholars 
had  he? 

Let  X  z=z  the  number  in  each  row ; 

then,        6  X  -|-  3  r=  the  whole  number  of  scholars ; 

also,        7  X  —  4z=z  the  whole  number  of  scholars. 

Hence,    7x  —  4  =  6x-\-3.   (ax.  7). 

10.  A  boy  being  sent  to  buy  a  certain  number  of  pounds  of 
meat,  found,  that  if  he  bought  pork,  which  was  9  cents  pei 
pound,  he  would  have  5  cents  left,  but  if  he  bought  beef,  which 
was  10  cents  per  pound,  he  would  want  5  cents.  How  many 
pounds  was  he  to  buy,  and  how  much  money  had  he  ? 

11.  Two  workmen  received  equal  wages  per  day;  but  if  the 
first  had  received  2s.  more,  and  the  second  2s.  less  per  day,  the 
first  would  have  earned  in  8  days  as  much  as  the  second  would 
in  12.     What  were  the  daily  wages  of  each  ? 

12.  A  and  B  began  trade  with  equal  stocks.  In  the  first  year 
A  gained  a  sum  equal  to  his  stock  and  $27  over ;  B  gained  a 
sum  equal  to  his  stock  and  $153  over.  The  amount  of  both 
their  gains  was  equal  to  five  times  the  stock  each  had  at  first. 
What  was  the  stock  with  which  each  began  1 

13.  A  man  is  40  years  old,  and  his  son  9;  in  how  many  years 
will  the  father  be  only  twice  as  old  as  the  son  ? 

14.  A  father  is  66  years  old  and  his  son  30 ;  how  many  years 
ago  was  the  father  three  times  as  old  as  his  son  ? 

15.  A  grazier  had  two  flocks  of  sheep,  each  containing  the 
same  number ;  from  one  of  these  he  sold  50,  and  from  the  other 
100,  ard  found  twice  as  many  remaining  in  the  one  as  in  the 
other.     How  many  did  each  flock  originally  contain  ? 

16.  A  courier,  who  traveled  80  miles  a  day,  had  been  gone 
one  day,  when  another  was  sent  from  the  same  place  to  overtake 
him.  In  what  time  will  the  second,  by  traveling  90  miles  per 
day,  overtake  the  first,  and  at  what  distance  from  the  starting- 
place  ? 

17.  A  gentleman  bought  a  horse  and  chaise;  for  the  chaise 
ke  gave  $75  more  than  for  the  horse,  and  three  times  the  price 


[V  EQUATIONS    OF    THE    FIRST    DEGREE.  1 1* 

of  the  horse,  diminished  by  $50,  was  equal  to  twice  the  price  of 
the  chaise.     Required  the  price  of  each. 

18.  When  wheat  was  worth  5s.  a  bushel  more  than  oats,  a 
farmer  gave  8  bushels  of  oats  and  8s.  in  money  for  4  bushels  of 
wheat.     What  were  wheat  and  oats  worth  per  bushel  ? 

19.  A  merchant,  engaging  in  trade,  during  the  first  year 
doubled  his  stock,  wanting  $500 ;  the  second  year  he  doubled 
the  stock  he  tljen  had,  wanting  $500;  and  so  continued  to 
double  his  stock  each  year,  wanting  $500 ;  until,  at  the  end  of 
the  fourth  year,  he  found  he  had  $500  more  than  eight  times  the 
stock  with  which  he  commenced.     What  was  his  stock  at  first? 

20.  Four  towns  are  situated  in  the  order  of  the  four  letters, 
A,  B,  C  and  D,  and  in  the  same  straight  line.  The  distance 
from  B  to  C  is  10  miles  less  than  twice  the  distance  from  A  to 
B ;  and  the  distance  from  C  to  D  is  20  miles  more  than  that 
from  B  to  C ;  moreover,  the  distance  from  A  to  B,  added  to  that 
from  B  to  C,  is  equal  to  the  distance  from  C  to  D  and  5  miles 
more.     What  is  the  whole  distance  from  A  to  D  ? 


SECTION    IV. 

EQUATION'S  OF  THE  FIRST  DEGREE,  CONTAINING  FRACTIONAIi  PARTS 
OF    SINGLE    TERMS. 

Art.  IS.  1.  A  merchant  sold  a  bag  of  coffee  for  $16,  which 

was  only  four  fifths  of  what  it  cost  him.    How  much  did  it  cost  t 

Let  X  z=.  the  number  of  dollars  it  cost. 

4x 
Then  four  fifths  of  x  may  be  written  f  x,  or  more  properly  — -, 

o 

wliich  may  be  read  either  four  fifths  of  x,  four  x  fifths,  one  fifth 

jf  four  X,  or  four  x  divided  by  five,  the  last  of  which  is  preferable 

4x 

Hence,     —  =  16.     Dividing  both  members  by  4,  we  have 

-  =  4 ;  if  one  fifth  of  x  is  equal  to  4,  the  who  e  of 
o 

I  will  be  five  times  as  much,  or,  z  =  $20,  Ans. 

2* 


18  EQUATIONS    OF    THE    FIRST    DEGREE.  IV 

Or,  we  might  first  multiply  by  5,  and  since  a  fraction  is  multi" 

4x 
plied  by  dividing  its  denominator,  we  have  —  or  4  x  =  80,  and 

X  =:  20,  as  before.     This  latter  method  is  generally  preferable  to 
the  former.  j 

2.  A  man  said  that  one  half  and  one  fourth  of  his  money 
amoimted  to  $75.     How  much  money  had  he? 

Let        X  =.  his  money. 

nr  nr 

Then     o  +  J  —    '^^'    Multiplying  by  2, 

X  +  -  =  150 ;  multiplying  this  by  2, 

2z  -\-x  =  300 ;  reducing, 

82;=:  300 ;  dividing  both  members  by  3, 
X  =  $100.     Ans. 

3.  In  a  certain  school,  one  half  of  the  boys  learn  Arithmetic , 
one  fourth,  French ;  one  eighth.  Grammar ;  one  sixteenth,  Al- 
gebra; and  10,  Geometry.  These  classes  constitute  the  whole 
school.     How  many  boys  does  the  school  contain  ? 

Suppose  x:=.  the  whole  number  of  scholars. 

Then,     x=    1+1  + 1+^  +  10.  (ax.  9). -Multiply  by  2, 
22:=    x  + 1 +  ^+1  +  20;  multiply  by  2, 
4x  =  2x  +  a:  +  |+^  +  40;  multiply  by  2, 

8x=4x+2x  +  x+^  +  80;  multiply  by  2, 

16x=i8x+4x  +  2x  +  x+160;  reducing, 
16x=  15x  +  160 ;  transposing  15  x  and  reducing, 
x=160.     Ans. 
Remark.     Although  it  is  generally  safest  to  multiply  by  the 
denominators  separately,  we  might,  in  this  question,  have  multi- 
plied Aie  first  equation  by  16,  the  least  common  multiple  of  the 


IV.  EQUATIONS    OF    THE    FIRST    DEGREE.  19 

denominators ;  or  we  might  have  reduced  all  the  unknown  terms 
to  a  common  denominator,  which  would  have  given  —x-  =—  -r-r 

nr 

4-10,  and  —  =  10,  consequently  2=  160. 

4.  A  man  found  that  he  had  spent  one  third  of  his  life  in 
Germany,  one  fourth  in  France,  two  fifths  in  England,  and  one 
year  in  the  United  States.  How  old  was  he,  and  how  many 
years  had  he  spent  in  the  first  three  countries  mentioned  ? 

5.  A  merchant,  on  settling  his  affairs,  found  that  he  owed  to 
one  man  ^,  to  another  ^,  and  to  a  third  ^-^  of  the  money  he  had 
on  hand;  and  that,  after  paying  them,  he  should  have  $3018 
left.  How  much  money  had  he,  and  how  much  did  he  owe  each 
of  the  three  creditors  ? 

6.  A  goldsmith  wished  to  make  a  mixture  of  gold,  silver  and 
copper,  so  that  2  ounces  more  than  one  third  of  the  whole  should 
be  gold,  8  ounces  more  than  one  fourth  of  the  whole,  silver,  and 
2  ounces  less  than  one  sixth  of  the  whole,  copper.  How  many 
ounces  in  the  whole  mixture,  and  how  many  of  each  kind  of 
metal  ? 

7.  A  man  left  his  estate  to  be  divided  between  his  wife  and 
his  three  sons,  in  the  following  manner,  viz :  the  wife  was  to 
have  $1000  less  than  one  third  of  the  whole  estate;  the  eldest 
son,  $2000  more  than  one  fifth  of  the  whole ;  the  second  son, 
$2000  more  than  one  sixth  of  the  whole ;  and  the  youngest  son, 
exactly  one  sixth  of  the  whole.  What  was  the  whole  estate,  and 
what  were  the  portions  of  the  several  heirs? 

8.  A  gentleman  had  spent  4  years  more  than  one  fourth  of 
his  life  with  his  parents  and  at  school,  12  years  less  than  three 
fifths  of  it  in  the  study  and  practice  of  his  profession,  and  had 
lived  in  retirement  20  years.     How  old  was  he  ? 

9  A's  age  is  to  B's  as  4  to  3,  and  if  twice  B's  age  be  added 
to  A  s,  the  sum  will  be  100  years.     Required  the  age  of  each. 

The  meaning  of  the  first  condition  is,  that  A's  age  is  f  of  B's, 
or  that  B's  is  f  of  A's. 

10  What  is  the  length  of  a  fish,  whose  head  is  3  inches  long 


20  EQUATIONS    OF    THE    F1R8T    DEGREE.  IV 

his  tail  f  the  length  of  his  body,  and  his  body  as  long  as  his  head 
and  tail  ? 

Let  X  z=z  the  length  of  the  body. 

11.  Three  fourths  of  a  certain  number  exceeds  five  ninths  of 
it  by  14.     What  is  that  number  1 

12.  A  person,  having  spent  one  half  of  his  money  and  one 
third  of  the  remainder,  had  $50  left.  How  much  had  he  at 
first? 

13.  Says  B  to  C,  lend  me  $200 ;  C  replies,  I  have  not  $200 
on  hand,  but  if  I  had  as  much  more  and  half  as  much  more  as  I 
now  have,  and  $12^,  I  should  have  $200.     How  much  had  he'' 

14.  Divide  60  cents  among  three  boys,  so  that  the  second 
shall  have  half  as  many  as  the  first,  and  the  third  10  more  than 
one  third  as  many  as  the  second. 

15.  A  man  wished  to  distribute  a  certain  number  of  apples 
amongst  his  four  children,  in  such  a  manner,  that  the  first  should 
have  one  third  of  the  whole ;  the  second,  three  fifths  as  many  as 
the  first ;  the  third,  two  thirds  as  many*  as  the  second ;  and  the 
fourth,  half  as  many  as  the  third  and  8  apples  more.  What  was 
the  whole  number,  and  how  many  would  each  child  receive? 

16.  A  gentleman  bought  two  horses  and  a  chaise ;  the  second 
horse  cost  once  and  a  half  as  much  as  the  first;  and  the  chaise 
cost  three  times  as  much  as  the  first  horse;  moreover  the  price 
of  both  horses  wanted  $50  to  be  equal  to  that  ^  of  the  chaise. 
What  was  the  cost  of  each  horse  and  of  the  chaise  ? 

17.  A  man  found,  that  he  expended  one  third  of  his  yearly 
income  for  board,  one  eighth  of  it  for  clothes,  and  one  twelfth  of 
it  for  other  purposes ;  and,  that  he  had  remaining  $550.  What 
was  his  income,  and  what  were  his  whole  expenses? 

18.  A  drover,  having  a  certain  number  of  sheep,  sold  one 
third  of  them  and  then  bought  60,  when  he  found  he  had  twice 
as  many  as  he  had  at  first.  What  was  his  first  number  of 
sheep  ? 

19.  A  gentleman  gave  to  three  persons  .£98.  The  second 
received  five  eighths  of  the  sum  given  to  the  first,  and  the  third, 
pne  fifth  as  much  as  the  secqnd.     What  did  each  receive  ? 


V  EQUATIONS    OP    THE    FIRST    DEGREE  21 

20.  A  person  set  out  on  a  journey,  and  went  one  se^  enth  of 
the  whole  distance  the  first  day,  one  fifth  the  second,  one  fourth 
the  third,  and  114  miles  the  fourth,  at  which  time  he  completed 
his  journey.  How  many  miles  did  he  travel  in  all,  and  how 
many  each  of  the  first  three  days  ? 


SECTION   V. 

EQl'ATIOJjrS     OF     THE     FIRST     DEGREE,      CONTAINING      FRACTIONAI, 
PARTS    OF    QUANTITIES    CONSISTING     OF     SEVERAL     TERMS. 

Art.  13.  1.  A  says  to  B,  I  am  6  years  older  than  you,  and 
two  thirds  of  my  age  is  equal  to  three  fourths  of  yours.  What  is 
the  age  of  each  ? 

Let  X  =.  B's  age , 

then,  a;  -j-  6  =  A^s  age. 

According  to  the  conditions  of  the  question,  three  fourths  of 
the  former  must  be  equal  to  two  thirds  of  the  latter.     One  third 

of  a: -|- 6  is  written    T^    ,  and  two  thirds  will  be  twice  as  much, 
o 

2x  +  12      „           3x      22  +  12      ,,,.,.      ^     ,    . 
or  ^ .     Hence,  -j-  = ^ .     Multiplymg  by  4, 

8x4-48         , „ 

o  X  = ;  multiplymg  this  by  3, 

9x=:8x-|-48;  transposing  and  reducing' 
X  =z  48  years,  B's  age.     x  -|-  6  z=:  54  years,  A's  age. 

Remark.  The  division  of  a  quantity  consisting  of  several 
terms,  as  x  -\-  6,  is  represented  by  placing  the  divisor  under  the 
dividend,  care  being  taken  to  extend  the  line  of  separation  under 
all  the  terms  of  the  quantity  to  be  divided, 

2.  A  man  bought  a  horse  and  saddle ;  for  the  horse  he  gave 
1*2:30  more  than  for  the  saddle ;  and  five  times  the  price  of  the 
saddle  was  equal  to  two  fifths  of  the  price  of  the  horse.  R&< 
quired  the  price  of  each. 


22  EQUATIONS    OP    THE    FIRST    DLGREE  V 

Let  %  ■=.  the  price  of  the  saddle ; 

then,  X  -f  230  =  the  price  of  the  horse. 

Hence,  according  to  the  conditions  of  the  question, 

5  X  = '- .     Multiplying  by  5, 

o 

25  X  =  2  2;  -{-  460 ;  transposing  and  reducing, 

23x  =  460;  dividing  by  23, 

X  =z  $20,  price  of  the  saddle, 

X  +  230  =:  $250,  price  of  the  horse. 

3.  A  father's  age  is  to  that  of  his  son  as  5  to  2,  and  the  dif 
ference  of  their  ages  is  30  years.     Required  their  ages. 

The  first  condition  signifies  that  the  father's  age  is  f  of  the 
son's,  or  that  the  son's  is  f  of  the  father's,  or  that  5  times  tho 
son's  is  equal  to  twice  the  father's. 

Suppose  X  =.  the  age  of  the  father ; 

then,       X  —  30  =  the  age  of  the  son. 

5a:_150      ,,,.,.       ^    ^ 
Hence,  x  = .     Multiplying  by  2, 

2  X  =  5  X  — 150 ;  transposing  and  reducing, 
—  3  X  =  —  150  ;  changing  the  signs, 

3  X  ==  150 ;  dividing  by  3, 

a:  z=  50  years,  father's  age ; 
X  —  30  =  20  years,  son's  age. 

4.  A  and  B  traded  together.  A  put  in  $100  more  than  B 
T  he  whole  stock  was  to  what  A  put  in  as  5  to  3.  How  much 
did  each  invest  in  trade? 

5.  A  man's  age,  when  he  was  married,  was  to  that  of  his  wife 
as  4  to  3;  but  after  they  had  been  married  10  years,  his  age  wag 
to  hers  as  5  to  4.  How  old  was  each  at  the  time  of  their  mar- 
riage T 

6.  A  man's  age,  at  the  time  of  his  marriage,  was  to  that  of  his 
wife  as  10  to  9;  but  if  they  had  been  married  10  years  sooner, 
his  age  would  have  been  to  hers  as  8  to  7.  What  were  their  re- 
upoctive  ages  at  the  time  of  marriage? 

7.  4  and  B  have  equal  sums  of  money;   but  if  B  gives  A 


V.  EQUATIONS    OF    THE    FIRST    DEGREE  23 

^€10,  ^  of  what  A  then  has,  will  be  equal  to  £  of  what  B  has  left 
How  much  money  has  each? 

&  Three  towns  are  situated  on  the  same  straight  road,  and  in 
the  order  of  the  letters  A,  B,  C.  The  distance  from  B  to  C  is 
20  miles  more  than  the  distance  from  A  to  B,  and  is  equal  to  | 
of  the  whole  distance  from  A  to  C.  What  are  the  distances 
from  A  to  B,  from  B  to  C,  and  from  A  to  C  ? 

9.  A  merchant  sold  three  packages  of  cloth ;  the  second  con- 
tamed  15,  and  the  third  30  yards  more  than  the  first;  moreover, 
the  third  contained  f  as  much  as  the  first  two.  How  many 
yards  were  there  in  each? 

10.  A  and  B  commence  trade  with  equal  stocks;  A  gains 
r^lO  per  year,  and  B  loses  «£5  per  year;  at  the  end  of  three 
years  B  has  only  f  as  much  property  as  A.  How  much  has  each 
at  first? 

11.  Two  boys,  standing  with  bows  and  arrows  on  the  bank  of 
a  river,  undertook  to  shoot  across  it ;  the  arrow  of  the  first  boy 
fell  10  yards  short  of  the  opposite  bank,  and  that  of  the  second 
fell  10  yards  beyond  it ;  now  it  was  found  that  the  first  boy  shot 
only  y\  as  far  as  the  second.  What  was  the  breadth  of  the 
river  ? 

12.  Two  men  have  equal  sums  of  money,  but  if  one  gives  the 
cither  $40,  the  former  will  have  only  |  as  much  as  the  latter. 
How  much  has  each? 

13.  A,  B  and  C  counting  their  money,  it  was  found  that  B 
had  $50  more  than  A  and  $75  less  than  C,  and  that  the  sum  of 
what  A  and  B  had,  was  f  of  the  sum  of  what  B  and  C  had 
How  much  money  had  each  ? 

14.  A  farmer,  having  a  certain  number  of  cows  and  twxs  aa 
many  sheep,  sold  15  cows  and  bought  5  sheep ;  he  then  found 
that  the  number  of  cows  was  to  the  number  of  sheep  is  3  to  13, 
Ilow  many  of  each  had  he  at  first  ? 

15.  A  man  engaged  to  work  a  year  for  $200  and  a  suil  of 
clothes ;  bat  falling  sick,  he  worked  only  5  months,  and  received 
$60  and  the  suit  of  clothes.  What  was  the  value  of  the  suit  of 
tlothes? 


24  eQUAlIONb    OF    THE    FIRST    DEGREE.  V 

16.  A  man  engaging  in  trade,  gained  the  first  year  6500,  but 
the  second  year  he  lost  -^  of  what  he  then  had ;  after  which  he 
found  that  his  stock  was  to  that  with  which  he  began  as  6  to  5. 
What  was  the  stock  with  which  he  commenced  1 

17.  A  grocer  bought  6  barrels  of  cider,  and  7  barrels  of  beer; 
he  gave  $2  a  barrel  more  for  the  beer  than  for  the  cider ;  and  | 
of  the  price  of  the  cider  was  equal  to  f  of  the  price  of  the  beer. 
What  was  the  price  of  each  per  barrel? 

18.  Two  numbers  are  to  each  other  as  9  to  10 ;  but  if  6  be 
added  to  each,  the  sums  will  be  as  10  to  11.  What  are  these 
numbers  1 

19.  A  man  built  two  pieces  of  wall,  one  jof  which  was  20  rods 
longer  than  the  other ;  for  the  shorter  he  was  to  have  $S  a  rod, 
Hnd  for  the  longer  $4  a  rod ;  now  the  whole  price  of  the  former 
was  to  that  of  the  latter  as  3  to  8.  What  was  the  length  of  each 
piece  1 

20.  A  gentleman  has  two  horses  and  one  chaise ;  now  if  the 
first  horse,  which  is  worth  $100,  be  harnessed,  he,  with  the 
chaise,  will  be  twice  the  value  of  the  second  horse ;  but  if  the 
second  horse  be  harnessed,  he,  with  the  chaise,  will  be  four  times 
*he  value  of  the  first  horse.  What  is  the  value  of  the  chaise  and 
of  the  second  horse  1 

21.  A  man  bought  a  horse,  and  afterwards  paid  $50  for  keep 
jng  him ;  he  then  sold  him  for  ^  of  what  he  had  already  cost  in- 
cluding the  keeping,  and  received  for  him  $20  more  than  he 
first  gave.     How  much  did  he  pay  for  him  at  first? 

22.  Two  cars  run  on  different  rail-roads;  the  speed  of  the 
second  is  2  miles  an  hour  greater  than  that  of  the  first ;  and  the 
distance  passed  over  by  the  first  in  8  hours,  is  f  of  that  passed 
over  by  the  second  in  9  hours.  What  is  the  speed  of  each  per 
hour  1 

23.  A  gentleman  started  on  a  journey  with  a  certain  sum  of 
racmey ;  after  having  had  $60  stolen  from  him,  he  expended  one 
third  of  what  he  had  left,  and  found  that  the  remaining  two  thirds 
wanted  $90  to  be  equal  to  the  sum  which  he  carried  from  home 
How  much  money  had  he  on  commencing  his  journe)  ? 


VI  ■  EQUATIONS    OF    THE    FIRST    DEC  REE.  25 

24.  A  man  having  a  gold  watch,  paid  $10  for  repairirg  it 
and  then  exchanged  it  for  two  silver  watches  of  equal  value,  and 
after  paying  $5  for  repairing  one  of  these,  he  found  that  it  had 
cost  him  $65.     What  was  the  value  of  the  gold  watch  at  first? 

25.  A  shepherd,  in  time  of  war,  was  plundered  by  a  party  of 
soldierg,  who  took  ^  of  his  flock  and  ^  of  a  sheep ;  another  party 
took  from  him  ^  of  what  he  had  left  and  ^  of  a  sheep ;  then  a 
ihird  party  took  ^  of  what  remained  and  ^  of  a  sheep ;  after 
vhich  he  had  but  34  sheep  left.     How  many  had  he  at  first  ? 


SECTION    VI. 

KQl/ATIONS    OF    THE  FIRST  DEGREE,  WHICH    REQUIRE   THE    SUBTRAC- 
TION   OF    QUANTITIES    COPfTAINING    NEGATIVE    TERMS. 

Art.  14:.  1.  A  and  B  commenced  business,  A  with  twice  as 
much  money  as  B ;  A  gained  .£20  and  B  lost  .£10  ;  then  the 
difference  between  A's  and  B's  money  was  .£70.  How  much 
did  each  begin  with? 

Suppose  we  knew,  that  B  had  .£40  and  A  <£80,  when  they  be- 
gan. Then,  after  A  had  gained  <£20,  he  would  have  80  -j-  20, 
or  £100;  and  B  having  lost  £10,  would  have  left  40—10  or 
£30 ;  now"  to  find  the  difference,  we  must  subtract  30  from  100, 
which  leaves  70.  But,  as  in  algebra  most  of  the  operations  can 
only  be  represented,  let  us  see  how  we  can  represent  the  prece- 
ding subtraction.  Instead  of  100  put  its  equivalent  80  -(-  20, 
and  instead  of  30,  its  equivalent  40  —  10 ;  our  object  is  to  sub 
tract  the  latter  from  the  former.  If  we  subtract  40  from  80  + 
2'0,  it  will  be  represented  thus,  80  -f-  20  —  40,  which  is  the  same 
as  GO;  but  we  wished  to  subtract  only  30  or  40  —  10  ;  we  hav« 
therefore  subtracted  too  much  by  10,  and  the  remainder  is  too 
small  by  10,  consequently  10  must  be  added  to  80-|-20  —  40 
which  then  becomes  80-|-20  —  40-j-  10  or  70.  Hence  we  see, 
that,  to  subtract  40  — 10,  we  must  change  the  -{-^0  to  — 40 
tnd  the  — 10  to  + 10 
3 


26  EQUATIONS    OF    THE    FIRST    DEGREE.  Vi 

Now  to  solve  the  question  ;  let     2;  =  B's  money  ; 
then  2xz=  A's  money. 

When  A  had  gained  .£20,  he  would  have  2x-|-20 ; 

and  B  having  lost  .£10,  he  would  have  x  —  10.     , 

Subtracting  B's  from  A's  gives  'Jix-]-20  —  2  -j-  10. 

For  X — 10  is  less  than  x  by  10,  and  if  we  subtract  x,  wt  sub 
(i  ict  too  much  by  10;  the  remainder  then,  after  x  has  been  sub- 
tracted, being  10  too  small,  10  must  be  added  to  correct  it 

Hence,         2  a; +20  — a;  +  10  r=  70.       Reducing    the    first 

member,         a;  -|-  30  r=  70 ;  transposing  and  reducnig, 

X  =  .£40,  B's  money ;  2x=:  .£80,  A's  money. 

2.  Divide  40  into  two  parts  such,  that  if  three  times  the  les* 
be  subtracted  from  twice  the  greater,  the  remainder  will  be  5. 
Suppose     X  =  the  greater  part ; 
then  40 — xz=z  the  less  part. 

For  if  the  greater  were  any  known  number  as  30,  for  example, 
the  less  would  be  the  remainder  of  40,  which  is  40  —  30  or  10 ; 
or  if  the  greater  were  28,  the  less  would  be  40  —  28  or  12.  So 
when  the  greater  is  represented  by  2,  the  less  will  be  40  —  t. 
Twice  the  greater  is  2x,  and  three  times  the  less  is  120  —  32 
which,  being  subtracted  from  22,  gives  22  — 120 -|- 3 2. 

Hence,  2  2  —  120  -|-  3  2  :=  5.     Transposing  and  reducing 
5xz=  125  ;  hence,  2  =  25,  the  greater  part, 
and  40  — 25  =:  15,  the  less. 

Art.  \5.  It  follows  from  the  preceding  questions  and  eTrpIana* 
lions,  that  any  quantity  is  subtracted  by  changing  the  signs  of 
all  its  terms f  and  writing  it  after  the  quantity  from  which  it  is 
to  be  subtracted. 

1  A  man  has  a  horse  and  chaise,  which  together  are  worth 
$400.  Now  if  the  value  of  the  chaise  be  subtracted  from  twice 
that  of  the  horse,  the  remainder  will  be  the  same,  as  if  tnree 
limes  the  value  of  the  horse  be  subtracted  from  twice  that  oj  the 
chaise.     Required  the  value  of  each. 

.    2.  A  vintner  has  two  equal  casks  full  of  wine ;  he  drawj  20 
.  .  llAcSvOut  of  one  and  40  out  of  the  other,  and  finds  the  differ 


VL  EQUATIONS    OF    THE    FIRST    DEGREE.  2? 

ence  between  the  number  of  gallons  remaining  in  the  two  casks 
equal  to  one  fourth  of  what  each  cask  contained  at  first  How 
many  gallons  does  each  cask  hold  ? 

3.  Divide  the  number  60  into  two  parts,  such  that  the  greater 
subtracted  from  50,  shall  be  equal  to  three  times  the  less  sul> 
tracted  from  90. 

4.  A  poulterer  had  a  certain  number  of  geese  and  twice  as 
many  turkeys  ;  after  having  sold  10  geese  and  bought  30  turkeys, 
he  found  that  if  he  subtracted  f  of  his  number  of  geese  from  his 
number  of  turkeys,  the  remainder  would  be  the  same,  as  if  he 
subtracted  y5_  of  his  number  of  turkeys  from  four  times  his  num- 
ber of  geese.     How  many  of  each  had  he  at  first  1 

Let      X  =.  the  number  of  geese ; . 

then  2xz=.  the  number  of  turkeys. 
After  selling  10  geese  and  buying  30  turkeys,  he  would  have 
X  —  10  geese  and  2  z  -|-  30  turkeys.     Then,  according  to  the 
conditions  of  the  question, 

_      ,   __       3x  — 30        .  ..       162;  +  240      -_  ,  .  .  . 

2  a;  4-30 =r4x  —  40 ^^ .     Muluplyrag 

o  lo 

by  5, 

10a;  +  150-^3x  +  30  =  20x  — 200  — ^^^^i^»  multiply. 

o 

ing  by  3, 
30Z  +  450  — 9a;  +  90  =  60a;  — 600— 16x  — 240;  transpos- 
ing  and  reducing, 

—  23  z  z=:  — 1380 ;  changing  the  signs, 
23  a;  =  1380 ;  dividing  by  23, 

X  bz  60,  the  number  of  geese ;  and 
22=  120,  the  number  of  turkeys. 
Observe  that,  after  the  equation  was  formed,  in  multiplying  by 
5,  3  z  was  changed  to  —  3  2,  and  —  30  to  +  30 ;  for,  the  sign 
—  precedmg  the  fraction,  belongs  to  the  whole  fraction,  and  not 
to  any  particular  part  of  it,  and  when  the  fraction  is  multiplied 
by  5,  the  numerator  is  to  be  subtracted ;  consequently  3  a:,  which 
is  supposed  to  have  the  sign  -f->  must  be  changed  to  —  3  a:,  and 
--30  to +30. 


28  EQUATIONS    OF    THE    FIRST    DEGREE.  VL 

Also,  in  multiplying  by  3,  both  the  terms  16  a;  and  240,  being 
iiffected  by  the  sign  -|-,  must  receive  the  sign  — .  If,  however, 
these  fractions  had  been  preceded  by  the  sign  -J-,  the  signs  of 
the  numerators  would  have  remained  unchanged.  The  same 
remarks  are  applicable  to  all  similar  cases. 

Care  must  be  taken  also,  when  fractions  are  preceded  by  the 
signs  -(-  and  — ,  to  make  these  signs  stand  e'en  with  the  lines 
ecparating  the  numerators  and  denominators 

5.  A  farmer  has  60  tons  of  hay  ;  of  this  he  sells  a  certain  por- 
tion, and  finds  that  ^  of  what  he  sells  subtracted  from  f  of  what 
he  retains,  gives  the  same  remainder,  as  f  of  what  he  retains  sub- 
tracted from  f  of  what  he  sells.     How  many  tons  does  he  sell  1 

6.  Two  men,  A  and  B,  set  out  on  a  journey,  each  with  the 
same  sum  of  money.  A  spends  $40,  and  B  $30 ;  then  f  of  A's 
money  subtracted  from  f  of  B's,  would  give  |  of  what  each  car- 
ried from  home.  How  much  money  had  each  on  commencing 
the  journey  ? 

7.  Divide  147  into  two  parts,  so  that  ^  of  the  less  subtracted 
from  the  greater,  shall  be  equal  to  -J  of  the  greater  subtracted 
from  the  less. 

8.  A  vintner  had  two  casks  of  wine,  each  containing  the  same 
quantity ;  from  the  first  he  drew  10  and  from  the  second  40  gal- 
lons ;  he  then  drew  from  the  first  ^  as  many  gallons  as  the  sec- 
ond contained  after  the  first  draught,  and  from  the  second  ^  as 
many  gallons  as  the  first  contained  after  the  first  draught,  and 
found  the  number  of  gallons  remaining  in  the  first  cask  to  the 
number  remaining  in  the  second  as  7  to  3.  How  many  gallons 
did  each  cask  hold? 

9.  A  market  woman  having  a  certain  number  of  eggs,  sold  30 
of  them,  and  found  that  f  of  what  she  had  left,  subtracted  from 
wliat  she  had  at  first,  would  leave  f  of  what  she  had  at  first 
Uow  many  had  she  before  she  sold  any? 

10.  A  man  having  a  lease  for  100  years,  on  being  asked  how 
much  of  it  had  already  transpired,  answered,  that  f  of  the  time 
past,  subtracted  from  |  of  the  time  to  come,  would  leave  the 
tame  remainder,  as  if  ^j  of  the  time  to  come  were  subtracted 


VTl.  MULTIPLICATION    OF    MONOMIALS  29 

from  f  of  the  time  past.     How  many  years  had  already  trans- 
pired ? 

11.  There  is  a  pole  consisting  of  three  parts;  the  middle  part 
is  4  feet  longer  than  the  lower  and  4  feet  shorter  than  the  upper 
part ;  moreover,  if  fV  of  the  upper  part  be  subtracted  from  §  of 
the  lower  part,  the  remainder  will  be  the  same,  as  if  ^  of  the 
middle  part  be  subtracted  from  |  of  the  upper  part.  What  is 
the  length  of  each  part  and  of  the  whole  pole? 

12.  If  a  certain  number  be  successively  subtracted  from  30 
and  52,  then  |-  of  the  first  remainder  be  taken  from  f  of  the  sec- 
ond, the  last  remainder  will  be  10.     What  is  that  number  ? 


SECTION    VII. 


MULTIPLICATION   OF   MONOMIALS. 


Art.  10.  It  may  be  remarked,  that  the  addition,  subtraction 
multiplication  and  division  of  algebraical  quantities,  cannot, 
strictly  speaking,  be  actually  performed,  in  the  same  sense  aa 
they  are  in  arithmetic,  but  are,  in  general,  merely  represented  ; 
these  representations,  however,  are  called  by  the  same  names  aa 
the  actual  operations  in  arithmetic. 

A  monomial,  or  simple  quantiti/,  consists  of  only  one  term 

(Art.  4) ;  as  a,  b  c  m,  or  — . 

A  binomial  is  a  quantity  consisting  of  two  terms,  ^  a-\-b^am 

a 
—  z  y ,  or  -  -j-  c  m. 

A  trinomial  is  a  quantity  consisting  of  three  terms,  as  a  +  ft 
1 —  cd. 

Polynomial  is  a  general  name  for  any  quantity  consisting  of 
leveral  terms. 

Moreover,  any  quantity  containing    more  than  one  term,  ii 
called  a  compound  quantity. 
3» 


30  MULTIPLICATION    OF    MONOMIALS.  VIl 

Art  17.  The  product  of  two  simple  quantities,  such  as  a  an.< 
6,  is  expressed,  either  by  writing  them  with  the  sign  of  multipli- 
cation between  them,  as  a  X  6  or  -a .  6,  or  by  writing  them  after 
each  other  without  any  sign,  as  ah.  The  last  form  is  most 
usual 

It  is  evidently  immaterial  in  what  order  the  letters  are  written ; 
for,  suppose  a  =  5,  and  6  =  7;  5  X  7  is  the  same  as  7X5; 
hence,  the  product  of  a  by  6  may  be  written  either  a  6  or  ha. 

In  like  manner,  the  product  of  a,  h  and  c  may  be  written  ahc^ 
achjb  a c,  he a^  ch a  Of  cah.  It  is  most  convenient  however 
to  write  them  in  the  order  of  the  alphabet. 

The  product  of  5 ah  by  2c d  might  be  written  5ah2cd; 
but,  as  the  order  of  the  factors  is  unimportant,  we  may  place  the 
numerical  factors  next  to  each  other,  thus,  5  X  2  a  6  c  6?,  or  per- 
forming the  multiplication  of  5  by  2,  we  have  10  ah  cd.  But 
we  could  not  write  5 2 a 6 c <f,  as  the  product  of  5 ah  and  2cd, 
because  the  value  of  a  figure  varies  according  to  its  place.  If 
we  wish  to  represent  the  multiplication  of  the  figures,  we  must 
separate  them  either  by  letters,  as  5  a  6  2  c  </,  or  by  the  sign  of 
multiplication,  sls  5  X  '^ a h  c d  or  5  . 2  ah  c d. 

The  same  result,  10 ah cdy  may  be  obtained  by  another  course 
of  reasoning ;  d  times  5 ah  is  5abdf  cd  times  5 ah  is  c  times 
as  much,  or  5 ah cd^  and  2cd  times  5 a 6  is  twice  as  much  as 
this  last,  or  10  ah  cd. 

By  a  similar  course  of  reasoning,  5ac,  4:bd  and  3»?n,  multi- 
plied together,  would  produce  60  ah  c  dm  n. 

We  see  from  the  preceding  examples,  that  the  product  of  iwo 
or  more  simple  quantities,  must  consist  of  the  product  of  the  co 
ifliicients,  and  all  the  letters  of  the  several  quantities. 

1.  Multiply  2 aw  by  76c.  6.  Multiply  13xy  by  12a. 

2.  Multiply  65  by  13c.  7.  Multiply  4p q  by  mn. 

3.  Multiply  3x1/  by  7 ah.  8.  Multiply  10 p  by  2am. 

4.  Multiply  Aar  by  6pq.  9.  Multiply  7 g-  by  3 m s. 

5.  Multiply  Sgh  by  17 ax        10.  Multiply  i^x  by  2 aq 

1 1.  Multiply  a  a  by  a  a  a. 


ni.  MULTIPLICATION    OF    MONOMIALS.  31 

The  product  in  the  last  question  would  be,  according  to  the 
principles  given  above,  aaaaa.  But  when  the  same  letter  en- 
ters into  a  quantity  several  times  as  a  factor,  instead  of  writing 
that  letter  so  many  times,  we  may  write  it  once  only,  and  place 
a  figure  a  little  elevated  at  the  right  of  it,  to  show  the  number  of 
times  it  is  contained  as  a  factor.  Thus,  a  a  is  written  a^ ;  a  a  a, 
a^y  and  aaaaa^  aP. 

A  product  must  always  contain  all  the  factors  both  of  the  mil- 
liplicand  and  multiplier.  In  the  present  case,  the  letter  a  is 
twice  a  factor  in  the  multiplicand,  and  three  times  in  the  multi- 
plier ;  therefore,  it  must  be  contained  five  times  as  a  factor  in 
the  product ;  that  is,  the  product  of  a^  and  c?  is  a^. 

In  like  manner,  the  product  of  3  a^  6^  and  4  a  6^  is  12  c?  h^ ; 
for,  each  letter  must  be  contained  as  a  factor  in  the  product,  as' 
many  times  as  it  is  in  both  multiplicand  and  multiplier.     Also 
the  product  of  4  a  6,  3  a  6^,  and  2a^  bnij  is  24  a^  b^  m. 

Art.  18.  This  figure,  placed  at  the  right  of  a  letter,  is  called 
the  index  or  exponent  of  that  letter,  and  aflfects  no  letter  except 
that  after  which  it  is  immediately  placed.  An  exponent  then 
shows  how  many  times  a  letter  is  a  factor  in  any  quantity. 

Letters  written  with  exponents  are  called  powers  of  those  let- 
ters; thus,  a^  is  called  the  second  power  of  a,  a^  the  third  power ^ 
a^  the  fourth  power ^  &c. ;  and,  for  the  sake  of  uniformity,  «, 
which  is  the  same  as  a^,  is  called  the  frst  power  of  a.  When  a 
quantity  is  written  without  any  exponent,  it  is  supposed  to  have 
1  for  its  exponent. 

In  some  treatises  a^  is  called  the  square,  and  a^  the  cube  of  a ; 
but  these  names  belong  to  geometry  rather  than  algebra.  Tha 
words,  square  and  cube,  however,  have  the  advantage  of  coa 
ciseness,  and  will  occasionally  be  used  in  this  work. 

Figures  also  may  be  written  with  exponents ;  thus, 

21  =  2. 

2^  or  2  .  2  =  4. 

23  or  2  .  2  .  2  =  8. 

2*  or  2  .  2  .  2  .  2  =  16. 


32  MULTIPLICATION    OP   Mi)NOMlALS.  VII 

The  expression  a^  h^y  if  2  be  put  instead  of  «,  and  3  instead 
©f  6,  becomes  2^  .  3^  or  2  . 2  .  2  . 3  .  3  =  72. 

Exponents  must  be  carefully  distinguished  from  coefficients ; 
for,  3  a  and  a?  have  very  different  significations.  Suppose  a  = 
10 ;  then  3  a  would  be  3  .  10  or  30,  but  a^  would  be  10  .  10 .  10 
or  1000. 

From  the  preceding  examples  and  observations,  we  derive  th« 
following 

RULE    FOR   THE    MULTIPLICATION    OF   MONOMIALS. 

Art.  19.  Multiply  the  coefficients  together^  and  write  after 
their  product  all  the  different  letters  of  the  several  quantilicSj 
giving  to  each  an  exponent  equal  to  the  sum  of  its  exponents  in 
all  the  quantities. 

I.  Multiply  a2  62  by  7  a3  64. 

In  this  question,  the  coefficient  of  the  multiplicand  is  1,  and 
that  of  the  multiplier  7,  the  product  of  which  is  7  . 1  or  7 ;  the 
8um  of  the  exponents  of  a  is  5,  and  the  sum  of  ^he  exponents  of 
fc  is  6 ;  hence,  the  answer  is  7  a^  h^. 

2.  Multiply    9  6c  by    6 a 6. 

3.  Multiply    7a3  66  by  43  a. 

4.  Multiply    2a2^2a;       by23a5m5. 

5.  Multiply  21  a  hy\^ amx. 

6.  Multiply  11  2;3  2^4  by      x^  y\ 

-    7.  Multiply    Qahcd       hy  I'ia^h'^  c^d^, 

8.  Multiply  73  m^  x*  by    2  m^  n^, 

9.  Multiply    5a263c4rf   hy^a^h^c^d. 
10.  Multiply    9pqx^y^   hy  IS p^ q x^ y"^ . 

II.  Multiply    2^3^3  2;       by    6a^h^n^x^ 

12.  Multiply  63  a  r  ^3  by    9a^rt*y. 

13.  Multiply  11  z2 3^2  by  Ux\yp^. 

14.  Multiply  I2amnp  by    3a^p^nm\ 

15.  Multiply  13  2:2  3^2  2  by    3xyz\ 

16.  Multiply  170  a  62  c  by       a^x. 

17    Find  the  product  of  10  a^,  3  a  6^,  and  7  rt  i  c. 
181    Find  the  product  of  3  c^  x,  2  x  y,  and  9p  x* 


yill  REDUCTION    OF    SIMILAR    TERMS. 

19.  Find  the  product  of  f  x^ ^2^  q^x y,  and  7pz^ yl 

20.  Find  the  product  of  f  a^  m^,  ^ax^  y^,  and  12  a  m  x^  y*. 


SECTION    VIll. 

REDUCTION    OF    SIMILAR    TERMS. 

Art.  SO.  When  several  monomials  are  connected  together  by 
the  signs  -\-  and  — -,  they  form  a  polynomial  or  compound  quan- 
tity.  But  such  polynomials  can  frequently  be  reduced  to  a 
smaller  number  of  terms.  This  can  be  done  when  some  of  the 
terms  composing  the  polynomial,  are  similar. 

Art.  SI.  Similar  terms  are  those  wjiich  are  entirely  alike  with 
regard  to  the  letters  and  exponents. 

N.  B.  In  determining  the  similarity  of  terms,  no  regard  is  to 
be  paid  to  the  numerical  coefficients,  or  to  the  order  of  the  let- 
ters. 

The  terms  3  a  6  and  5  a  6  are  similar ;  so  are  6  a^  b^  and  9  a^  b^. 
But  2  a^  b^  and  4  a^  b^  are  not  similar,  because  the  exponents  of 
a,  as  well  as  those  of  b,  are  different  in  the  two  quantities. 

Suppose  we  have  the  polynomial  6a^b  —  2  ?n  n -\- i  a^  b  -\-' 
6mn.  Here  6a^b  and  4a^b  are  similar,  and  it  is  evident  that 
6  times  and  4  times  the  same  quantity  make  10  times  that  quan 

y;  hence,  6  a^b-\-4:a^b  is  10  a^  6  ;  also,  — 2mn-{-(Jmn^ 
w  6mn  —  2w»,  is  +4mw;  the  given  polynomial,  therefore, 
becomes  10  a^  b -\- 4:  m  n. 

Again,  take  the  polynomial,  2462c3  —  Sabc^  -{-2b'^c^ — 5pq 
^Qabc^—2pq-^^m^z.  Here,  24 6^ c3  +  2 ft^  c3  —  26 52 c3 ; 
—  3a6c2  —  Gab c^  =.  —  9abc^',  and  —  ^ P  q- —  2p q:=.  — 
7p5;  hence,  the  given  quantity  becomes  ^Qh^c^- — 9a6c2  — 
7j9  q  —  37»2«. 

Suppose  the  following  polynomial  given,  viz  :  'ia^bc^ — ^a^bc 
+  12  a2  6  c2  —  4  a3  5  c  +  12  a3  6  c  +  5  a2  6  c2  +  3  a3  6  c  -^ 
10  a^bc^  —  a^b  c^.  First,  collect  the  positive  terms  of  one  kind  * 
2a2&c2  +  12a26c2  +  5a26c2=19a26c2;   then  collect  the 


34  REDUCTION    OF    SIMILAR   TERMS.  VIll 

negati  re  terms  of  the  same  kind ;  — 10  a^  b  c^  —  a^  6  .r^  j=  — 
11  a^  b  c^;  the  five  terms,  therefore,  are  the  same  as  19  cfibc^  — 
{{  a^b c^,  or  8 a^ 6 c^.  In  like  manner,  the  positive  and  negative 
♦erms  of  the  other  kind  being  separately  collected,  give  \5a^bc 
— 8a^bc,  or  7 a^ be.  The  polynomial  therefore  becomes 8a^bc^ 
+  7aHc. 

Sometimes  the  sum  of  the  negative  terms  exceeds  that  of  the 
similar  positive  terms ;  in  such  cases,  we  must  take  the  difference 
between  the  two  sums  and  give  it  the  sign  — . 

Suppose  we  have  m  +  Sab^  ^I3ab^  +  10  ab^  —  \2  ab^ ; 
collecting  the  similar  positive  and  negative  terms  separately,  we 
have  m-\-18ab^  —  25  a  6^ .  t^jg  jg  (\^q  game  as  w  -}-  18  a  6^  — • 
18  a  b^— 7  a  b^.  But  +  18  «  62  and  —I8ab^  destroy  each 
other,  and  the  result  is  m  —  7  ab^. 

Art.  32.  Hence  we  derive  the  following 

RUL.K    FOR   THE    REDUCTIONT    OF    SIMILAR    TERMS. 

Unite  all  the  similar  terms  of  one  kind  affected  with  the  sign 
-}-,  by  adding  their  coefficients  and  writing  the  sum  before  the 
common  literal  quantity ;  unite,  in  like  manner,  the  similar  terms 
of  the  same  kind  affected  with  the  sign  — ;  then  take  the  differ* 
ence  between  these  two  sums,  and  give  the  result  the  sign  of  the 
greater  quantity. 

Remark.  The  learner  will  be  less  liable  to  error,  if  he  take 
the  precaution  to  mark  the  terms,  as  he  reduces  them. 

Reduce  the  following  polynomials  to  the  least  number  of  terms. 

1.    6a^b  —  8a^b  —  9a^b+l5a^b  —  ab^. 
'    2     7a6c2— a6c2  — 7a6c2— 8a6c2+12a6c2. 

3.  16a6c3  +  5rt63c  +  7a36c  — 10a6c3  — 7a36.c  — 4nin 
^4a63c_6a63^2a6c3. 

4.  12p3cr+16m2n2— .c  +  4cp3r  +  3c  +  75^*  — 17;)3cf 
-  12  m2  n2  — 2  c  +  3m2  n^. 

5.  67«wr+llp3y__i7;re„r  +  36c  — 22;j3y  _^1Sot7i 
f  -  37>3  q  —  5mnr  +  76c. 

6.  22x2  — y3_|_32;2_|_822  — 7x2  +  7n  +  7y3_i6y3_f.3;„ 

\   7»i  +  2;2-f  2y3__6w». 


QC.  ADDITION.  36' 

v^mp-^^amp^  —  Samp^ —  12 a^mp  -\-  10 a^m p. 
8.    abc^  +  7mn^p^r'^  +  Qabc^  +  9ab^c-{-l2m^n  +  2abc» 

a62c-[-a6c2  — 3w2»2p3^4_3a6c3--2a6c2. 


SECTION    IX. 


ADDITION. 


Art.  33.  The  addition  of  positive  monomials  or  simple  quati" 
titles  J  consists  merely  in  writing  them  after  each  other,  and  giv- 
ing to  each  of  them  the  sign  4-»  except  the  first,  which  is  also 
supposed  to  have  the  sign  -f-.  Thus,  to  express  the  addition  of 
a  and  6,  we  write  a-\-b  or  b-^-a.  Also,  to  signify  the  addition 
of  a,  6,  c  and  d^  we  write  a  -|"  ^  H"  ^  H~  ^*  ^^  ^^^®  manner,  the 
addition  of  a  6,  3zy,  and  4mn,  is  expressed  thus,  ab-\-^xy  -\- 
Amn  The  order  of  the  terms,  as  was  observed  in  Art.  O,  ia 
unimportant. 

Art.  241.  If  it  were  required  to  add  together  the  polynomials, 
a  -f-  6  -|-  c,  and  m  -f-  w,  in  which  all  the  terms  are  affected  with 
the  sign  -{-,  the  process  would  evidently  be  the  same,  as  if  it  were 
required  to  add  together  the  separate  terms  of  which  these  poly- 
nomials are  composed ;  that  is,  we  should  write  them  after  each 
other,  giving  the  sign  -(-  to  every  term  except  the  first ;  and  the 

sum  would  bea-|-6-f-'^~l~'"~h'** 

But  if  some  of  the  terms  in  the  polynomials  to  be  added,  have 
the  sign  — ,  they  must  retain  the  same  sign  in  the  sum.  Take 
an  example  in  arithmetical  numbers.  Let  it  be  proposed  to  add 
10  —  3  to  .12 ;  10  —  3  is  7 ;  we  wish  then  to  add  7  to  12.  But, 
if  we  first  add  10,  which  \s  expressed  thus,  12  -\-  10,  the  sum  is 
too  great  by  3 ;  therefore,  after  having  added  10,  we  must  sub- 
tract 3,  and  the  true  sum  is  12  +  10  —  3  or  19. 

Let  us  no\^  add  b  —  c  to  a.  First  add  6  to  a,  and  we  have  a 
4-  6  •  this  is  too  great  by  c,  because  the  quantity  b  —  c,  which 


d6  ADDITION.  IX 

was  to  be  added,  is  less  than  h  by  the  quantity  c ;  therefore,  aftei 
having  added  6,  we  must  subtract  c,  and  the  true  result  is  a  -j- 

We  see,  in  these  instances,  that  we  have  merely  written  the 
terms  after  each  other  without  any  change  in  their  signs ;  and  the 
reasoning  used  in  explanation  of  the  process,  is  applicable  to  the 
addition  of  all  polynomials,  in  which  some  of  the  terms  are  af- 
fected with  the  sign  — . 

The  sum  of  the  polynomials  c^h-\-^c  —  4a  and  Vic-\Sa^h 
—  3a  isa26  +  3c  — 4a+12c  +  8a26  — 3a.  But  this  sum 
contains  similar  terms,  which  may  be  reduced,  according  to  the 
principles  given  in  Art.  2S8.  This  reduction  being  made,  the 
sum,  in  its  simplified  form,  becomes  9 a^ 6  -|-  15 c  —  la. 

Art.  25.  From  what  has  been  said  above,  we  deduce  the  fol 
lowing 

BULK    FOR   THE    ADDITION    OF    ALGEBRAIC    QUANTITIES. 

Write  the  several  quantities  one  after  another ^  giving  to  each 
term  its  proper  sign,  and  then  reduce  the  similar  terms. 

Observe,  that  those  terms  which  have  no  sign,  are  supposed  to 
have  the  sign  -j- 

1.  Add  together  4  a,  6  6,  7  c,  9  a, 

3a  +  6,  6c,  4</, 
and  4a  +  3c  —  4&. 

2.  Add  together  3  a  6  —  Acd,  m^n, 

9a6+8c</,  3»i2w  — w, 
and  4  m. 

3.  Add  together  3a6-|-4c£? —  m^, 

4  ctn  —  7  cd~\-Saby 
12a6  +  8c(/+6w2. 

4.  Add  together  m^, 

flS  62  ^  6  m2  —  6  m  w, 
4a252__i2»iw  +  8m«, 
4xy  — 7m2  +  36  +  8a«69. 
6    Add  together  11  bc-\- 4^ ad — 8ac-|-5crf, 
8ac  +  76c  —  2ad-\-4Lmn^ 


X  suba^iaction.  87 

2cd  —  ^ab-\-5ac-\'  an^ 
dan  —  ^bc  —  ^ad  +  Scii 
5,  A  (Id  together  5a  +  46  —  3  c, 

3a-— 126, 

7c  — 10<f  — 4, 
16  — 3c  +  5. 
7   Add  together  3  6  —  a  —  c^-U5d, 
Qc—^f—d, 
3a_26  — 3c  +  27c, 
3c  — 7/+56  — 8c, 
17c  — 66  — 7a, 
11/— 5e  +  9e/  +  ^-3a, 
6c  — 5c  — 2rf  — 9/. 

8.  Add  together  4  a2  6  +  3  c3  c?  —  9  w2  M, 

4m2n  +  a62  +  5c3<f  +  7a2  6, 
6m2w— 5c3rf  +  4mw2  — 8a62, 
7»iw2  +  6c3</— 5m2»  — 6a2  6, 
7c3rf— 10a6  — 8m2w  — lOe/4, 
12a2  6  — 6a62  +  2c3rf  +  »i». 

9.  Add  together  2a62  +  3ac2  —  8c22  +  962  2._8^y2^ 

5a3_4a52__7fta;2^ft2a;__4^3,2_i5^y^ 

5  A:.y— ^3,2^112;  ^1453_22ac2  —  10z« 
19ac2  — 8  622;  +  9x2  +  6Ay  +  2A:y2, 
2a62+7i)3^2__ioA;y  +  3a3  +  2z. 


SECTION    X. 


SUBTRACTION. 


Art.  26.  We  have  already  seen,  that  a  simple  quantity  is  sub- 
tracted, by  giving  it  the  sign  — ;  thus,  to  subtract  6  from  a,  we 
write  a  —  6. 

We  are  now  to  show  how  to  subtract  polynomials.  If  it  were 
required  to  subtract  7-1-3  from  12,  it  is  evident  that  7  and  3 
4 


^ 


SUBTRACTION. 


must  both  be  subtracted,  which  is  expressed  thus,  12  —  7 — 3 
In  like  manner,  ifb-\-c  is  to  be  subtracted  from  a,  b  and  c  musf* 
both  be  subtracted,  thus,  a  —  b  —  c. 

But  if  some  of.the  terms  in  the  polynomial  to  be  subtracted, 
have  the  sign — ,  the  signs  of  these  terms  must  be  changed  to-j- 
Suppose  it  were  required  to  subtract  7 — 5  from  10;  7 — 5  is  2, 
and  2  from  10  leaves  8.  Now  if  we  subtract  7  from  10,  which 
is  represented  thus,  10 — 7,  we  subtract  too  niuch  by  5,  and  the 
remainder  3  is  too  small  by  5 ;  consequently,  after  having  sub- 
tracted 7,  we  must  add  5,  and  the  true  result  is  10  —  7-j-5,  or  8. 

Now  let  us  subtract  b — c  from  a.  First  subtract  6,  and  we 
obtain  a — b;  but  b  is  greater  that  b  —  c  by  c;  therefore,  as  we 
have  subtracted  too  much  by  c,  the  remainder  is  too  small  by  c ; 
we  must,  consequently,  add  c  to  a — 6,  and  we  have  a — b-^-c 
for  the  true  result. 

The  same  reasoning  is  applicable  to  the  subtraction  of  all 
polynomials  containing  negative  terms. 

Art.  3T.  Hence,  we  deduce  the  following 

RULE    FOR   THE    SUBTRACTION    OF    ALGEBRAIC    QUANTITIES. 

Change  the  signs  of  all  the  terms  in  the  quantity  to  be  sub- 
traded  J  and  write  it  after  that  from  which  it  is  to  be  subtracted; 
then  reduce  the  similar  terms. 

1 .  From  8  a  +  4  &,  subtract  3  « — 26. 

Changing  the  signs  of  the  latter  quantity,  and  writing  it  after 
the  former,  we  have  8a-j-46 — 3a-|-26,  which  reduced,  be- 
tomes  5  a  +  <)  6    Ans. 

2.  From4a6 — 36c,  subtract  2a6— 66c.   Ans.  2a6-f-36c 

3.  From  4a6  —  "Sc^-^-bc,  subtract  ab —  c*  —  2  6  c. 

4    From  5ac  —  8a6-j-96c  —  4am,  subtract  8am  —  2ah 
|-  II  ac  —  7  cd. 
5.  From  3m— Sx  — 7/,  subtract  3  rf— 6^.-2  2— 6/ -f  8 

Art.  38.  Sometimes  it  is  conveiiient  to  ro'^reseni  the  subtrao* 
ti.on  of  polynomials,  without  actually  pfr.'ormlrg  the  cperatloii, 


X  SUBTRACTION.  l3i» 

This  is  done  by  enclosing  the  quantity  within  brackets  or  a  f  e'cw- 
thesis  J  and  prefixing  the  sign  — .    Thus,  Qah  —  ^c-\-  d —  (3a6 

—  4  c  -[-  2  6?)  signifies  that  3a6  —  4c-|-2c?istobe  subtracted 
from  Q  ab  —  3c-|-6?.  Performing  the  subtraction  indicated  and 
reducing,  remembering  that  3  a  6  within  the  parenthesis  has  the 
-j-  sign,  we  obtain  3  a  6  -|-  c  —  d. 

According  to  this  principle,  a  polynomial  may  be  made  to  un- 
dergo various  transformations. 

For  example,  3  a  6  —  a  —  b  is  the  same  as  3  a  —  {a-\-b)\  for, 
when  the  subtraction,  indicated  in  the  latter  expression,  is  per- 
formed, it  becomes  3  «  6  —  a  —  b. 

In  like  manner,  la?  —  Q  a^b  —  4  6^  c  -j~  6  6^  is  equivalent  to 
7  a3  —  8  a2  6  -_  (4  62  c  —  6  63). 

Let  the  learner  perform  the  subtraction  indicated  in  the  fol- 
owing  examples. 

1.  27a2x  — 26c  +  4x2_j-3aa;  — (9a2x  +  4  6c  — 6x2  —  6 
+  4  a  X  +  6). 

2.  28ox3  — 16a2x2  +  25a3x-— 13a4  — (18ax3  +  20a2x2 
-_24a3a:__7«4). 

3.  30«6  — 6  6c2  +  2  62_45__(4a6-[.i26c2  — 2462_. 

4.  8a2xy  — 5  6x2y-f-17cxy2«_9^5_(a2a;y +  36x25, 

—  13cx3,2_j_20y5). 

5.  63  x2  3,2  _j_  24  X  y  +  62  c2— .  c  d3  +  5  c3  cf  —  (45  x2y2  _ 
24  X  y  —  3  62  c2  — 2  c  c?3  _j-  4  c3  rf  —  ;re). 

Art.  39.  The  reverse  of  the  process  in  Art.  28  is  sometimes 
very  useful,  viz :  putting  within  a  parenthesis  part  of  a  polyno- 
mial, and  placing  the  negative  sign  before  the  parenthesis.  To 
do  this,  it  is  only  necessary  to  change  the  signs  of  all  the  terms, 
placed  within  the  parenthesis. 
.     Thus, a  +  6  —  c  =  a  —  ( — b-\-c)=:a — (c— 6);  also,4a6c 

—  6c2^  +  m2— 7pg  =  4a6c  — (6c2rf  — m2  +  7p7). 

Let  the  learner  throw  the  last  two  terms  of  each  of  the  fol- 
lowing quantities  into  a  parenthesis,  preceded  by  the  negative 

BlgU 


40  MULTIPLICATION    OF    POLYNOMJALS.  XI 

1.  am  —  hc-\-dm. 

2.  ah  c  —  c?2  —  3»ty  +  a;y. 

3.  a  -f-  ^  —  ^  —  ^• 

5.  4m2  +  12m7i  +  9w3_2TO  —  3» 

6.  iitth  -\-l[  am  —  x^-\-y^, 

7.  7  m  »  -]-  x^  — pq-\-y^* 


SECTION    XI. 

MULTIPLICATION    OF    POLYNOMIALS. 

Art.  30.  Multiply  10  +  3  by  7.     It  is  manifest,  that  10  and 
3  must  both  be  multiplied  by  7,  and  the  products  added. 
Operation, 
10+3 

_7 

70  +  21  =r  91  =  13,  7. 
So,  to  multiply  a  +  6  by  c,  each  of  the  terms,  a  and  6,  must 
be  multiplied  by  c,  and  the  products  added. 
Operation, 
a    +    h 


ac-\-h  c. 
If  there  are  several  terms  in  both  multiplicand  and  multiplier, 
all  the  terms  of  the  former  must  be  multiplied  by  each  term  of 
he  latter. 

Multiply  8  +  3  by  7  +  5.     Here  we  must  take  8  +  3  seven 
imes  and  five  times,  and  then  add  the  products. 
Operation. 
8  +  3 

Z  +  5 

66  +  21  +  40  +  15=133 


XL  MULTIPLICATION     )F    POLYNOMIALS.  4 

If  both  quantities  be  reduced  before  multiplying,  and  then 
their  product  taken,  the  result  will  be  the  same;  thus,  11.12 
=  132. 

In  like  manner,  if  a  -(-  6  is  to  be  multiplied  by  c-{-dy  we  mus 
multiply  a-\-b  by  c  and  then  by  d,  and  take  the  sum  of  the  pro- 
•^ucts. 

OpercUion. 
a  +  b 

c  +  d 

ac-^-bc  -j-'a  d-\-bd. 

Let  us  suppose  now  that  the  multiplicand  contains  a  negative 
term. 

Multiply  10  — 6  by  3.  The  multiplicand,  10  —  6,  is  4,  and  3 
times  4  is  12.  But,  if  we  multiply  10  by  3,  the  product  30  is  too 
great  by  3  times  6  or  18,  which  must  therefore  be  subtracted 
from  30 ;  the  true  product  then  is  30  —  18  or  12. 

Operation, 
.10  —  6 

3 

30  —  18=12. 

So,  to  multiply  a  —  b  by  c;  if  we  first  multiply  a  by  c,  the 
product  a  c  is  too  great  by  c  times  6  or  6  c,  which  must  therefore 
be  subtracted  from  a c ;  the  true  result  then  is  a c  —  be. 

Operation, 
a  —  b 
c 


ac  —  be. 

The  term  —  6  c  in  the  product,  shows,  ahat  when  a  negative 
term,  as  —  6,  is  multiplied  by  a  positive  term,  as  +  c,  the  prO" 
duct  must  be  negative. 

Now  let  both  multiplicand  and  multiplier  contain  negative 
ferms. 

Multipl}  18  —  3  by  12  —  5.  If  we  reduce  both  numbers, 
4* 


12  MULTIPLICATION    OF    POLYNOMIALS.  Xi. 

ana  then  multiply  them  together,  the  product  is  15  .  7  or  105. 
But  to  perform  the  operation  without  reducing,  we  first  multiply 
18  —  3  by  ]2,  which  gives  216  —  36;  but,  as  we  wished  to  take 
18  —  3  only  12  —  5  or  7  times,  we  have  taken  it  5  times  too 
many  times;  we  must,  therefore,  subtract  5  times  18  — 3  or  90 
—  15  from  216  —  36,  which  gives  216  —  36  —  90  +  15.  See 
^Tt.  27. 

Operation, 
18  —  3 
12  —  5 


216  —  36  —  90  +  15  =  105. 

In  a  similar  manner,  to  multiply  a  —  6  by  c  —  d,  we  first  take 
c  times  a  —  6,  which,  as  we  have  already  seen,  is  ac  —  be;  but 
as  we  wished  to  take  a  —  b  only  c  —  d  times,  we  have  taken  it  d 
times  too  many  times.  Now  d  times  a  —  b,  d  being  considered 
as  positive,  is  ad — bd;  this  then  must  be  subtracted  from  ac 
—  6c,  and  we  have  ac  —  be  —  ad-\-bd  hr  the  true  product  of 
a  —  6  by  c  —  d. 

Operation 
a  —  b 
c  —  d 
ac  —  be  —  ad-\-bd. 

We  see  that  the  term  —  ad  \s  produced  by  multiplying  -f- a 
by  — d;  hence,  if  a  positive  term  be  multiplied  by  a  negative, 
the  product  must  be  negative.  Moreover,  the  term  -\-bd\s  pro- 
duced by  multiplying  — b  by  — d\  therefore,  if  two  negative 
terms  are  multiplied  together,  the  product  must  be  positive. 

Art.  31.  From  the  preceding  explanations,  we  derive  the  fol- 
owing 

RTTLE    FOR    THE   MULTIPLICATION   OF   POLTWOMIALS. 

1.  Multiply  all  the  terms  of  the  multiplicand  by  each  term  of 
the  multiplier  separately,  according  to  the  rule  for  the  multiplica 
Hon  of  simple  quantiti^.s. 


XI.  MULTIPLICATION    OF    POLYNOMIALS.  43 

2.  With  regard  to  the  signs,  observe,  that  if  the  two  terms  to 
be  multiplied  together,  have  the  same  sign,  either  both  -\-  or  both 
— ,  the  product  must  have  the  sign  -{- ;  but  if  one  term  has  the 
sign  -f-  and  the  other  the  sign  — ,  the  product  must  have  the 
sign  — . 

3.  Add  together  the  several  partial  products,  reducing  terms 
which  are  similar. 

1.  Multiply  2a6  +  6c  +  32;y 

hy^ab  +  'ibc 

Qa^b'^  +  Zab'^c+^abxy  \    Partial 

-j-4q6Qc-|-26^cQ  +  66czy  |  products. 
Qa^b'^  +  1  ab^c+'ilb^c^+^abxy  +  Qbcxy', 
which  is  the  result  reduced. 

Remark,  It  is  convenient,  in  order  to  facilitate  the  reduc- 
tion, to  place  similar  terms,  in  the  partial  products,  under  each 
other. 

2.  Multiply 

4a3_  ^a^h—  8a&2^253^by 

2a2—    3q  6—  46^ 

8a5  — 10a46_i6a362-|_   4^253  )    p^    # 
—  I2a46  +  I5a3  69_|.24a263_   Qah^  )    ^*"  , 
—  16a3  62.|_20a2fe3  4,32a64__3^^5)  products. 

Ba'i  — 220^6  — 17a3  62^48a263_j_26a64__865.    Result  re- 
duced. 

3.  Multiply 

a^  +  aH  +  a^b^  +  a^b^  +  al^-^b^,}^^) 
a  — b 

a6  _|.  «5  6  _]_  ^4  ^2  _|_  ^3  ^,3  _j.  a2  64  ^  a  65  I    Partial 

-,q5;,_a4fe2_a3fe3_fl2M__fl55_56  ]  products 
a6  —  56^     Result  reduced. 

4.  Multiply  2  o  —  3  by  a  +  4. 

5.  Multiply  a2  +  2  a  6  +  2  62  by  a2  _  2  a  6  +  2  62. 

6.  Multiply  7a6  —  3ac2  +  4c2(Z2  by  4«c2  —  3a6  — 
2c2r/2. 

7.  Multiply  14a3c_66c  +  12xy2_c2  by  Qa^c+^ht 
—  2x2y_[.3c2. 


14  MULTIPLICATION    OF    POLYNOMIALS.  XI 

8.  Multiply  3  a4  6  _  4  a3  52  _|_  6  a9  63 _  «  64  by  a^  —  2ah 
4-62. 

9.  Multiply  1463  +  2862c--76c2  +  c3by  362— 66c— 2c9. 
Art.  33.    Sometimes   the   multiplication   of   polynomials  ia 

indicated^  without  being  actually  performed.  This  is  done  by 
placing  a  horizontal  straight  line,  called  a  vinculum,  over  each 
of  the  polynomials,  and  writing  them  after  each  other  with  the 
sign  of  multiplication  between  them ;  or,  by  enclosing  each  in  a 
parenthesis,  and  writing  them  after  each  other,  eitker  with  or 
without  the  sign  of  multiplication.  Thus,  each  of  the  expres- 
sions, a  -["  *  X  »*  —  w,  a-{-  b  .  m  —  n,  (a  +  6)  X  (»» —  w), 
{a-\-b)  .  (m  —  w),  and  (a  -{-b)  (m  —  n),  represents  the  multi- 
cation  oi  a-^-bhy  m  —  n. 

The  last  mode  of  representation  is  generally  preferred ;  but 
we  must  be  careful  to  include  each  polynomial  in  a  parenthesis ; 
for,  (a-\-b)m  —  n  indicates  that  a-\-b\s  multiplied  by  m  only, 
and  that  n  is  subtracted  from  the  product. 

In  like  manner,  {a-{-b)  (m  —  n)  (x-j-y)  indicates  that  the 
first  polynomial  is  multiplied  by  the  second,  and  that  product  by 
the  third. 

Let  the  learner  perform  the  operations  indicated  in  the  follow 
ing  examples. 

1.  (a2  +  2a6  — c2)  (a  — 6). 

2.  (a2  +  2a2j_262)  (a2_2a26  +  262). 

3.  (m3  +  m2  w  +  m  w2  +  7i3)  (m  —  n)  (m*  +  w^). 

4.  (a2_^  62_[_  c2)  (6  +  c)  +  (a2_  J2_c2)  (6  — c). 

In  the  last  example,  the  first  polynomial  is  to  be  multiplied  by 
the  second,  and  the  third  by  the  fourth,  and  then  the  products 
are  to  be  added. 

5.  (3x2  +  6xy  +  33^2)(x_y)  +  (4x2_6xy  — 3y2)  x 

6.  (x^a  —  b)  (4ac  — 2)  +  (2x  +  2a  +  36)(3ac  +  2). 

7.  (a2_^2a6  +  62)  {c  +  d) —(a^ —2ab  +  b^)  {c^d). 

In  this  last  example,  the  product  of  the  last  two  polynomials  ii 
10  be  sub  .racted  from  the  product  of  the  first  two. 


KL  MULTIPLICATION    OF    POLYNOMIALS.  45 

a  (4x2  — 4a;y  +  c2)  (a  +  6)  — (a2;„  +  a:y)  (c2_^)  x 
(a  +  x). 

9.  62(l36c2x  — 7c3x2)_c2(62c22— 18c3a;2). 

10.  (112  — 4x  +  a2)  (6  +  4)+a2(io  +  66c  — 7c2). 

Art.  33.  The  following  cases  of  multiplication  deserve  par 
licular  attention,  on  account  of  the  practical  application,  which 
w  ill  hereafter  be  made  of  the  results. 

Let  a  and  h  represent  any  two  quantities ;  their  sum  is  a  -j-  i, 
and  their  difference  a  —  6.     Multiply  a-\-bhy  a —  6. 
Operation, 
a-\-h 

a^  +  ah 
^ab  —  b^ 


a^  —  b^ 

The  product,  a^  —  b^j  is  the  difference  between  the  second 
Dower  of  a  and  the  second  power  of  6.     Hence, 

The  sum  of  two  quantities  multiplied  by  their  difference,  gives 
the  difference  of  the  second  powers  of  those  quantities. 

Suppose,  for  example,  two  numbers,  10  and  4 ;  then,  (10  +  4)  X 
(10  —  4)  =  100—16  =  84. 

In  like  manner,  (3  a  +  4  6)  (3  a  —  4  6)  =  9  a2  —  16  62. 

Art.  34.  When  a  polynomial  is  multiplied  by  itself,  the  re- 
sult is  called  the  second  power  of  that  polynomial,  and  when  it  !a 
multiplied  twice  by  itself,  the  result  is  called  the  third  power. 
Find  the  second  power  of  the  binomial  a-^  b. 
Operation, 
a  +  b 
a-\-b 
a'^+ab 
-f  «6  +  62 


a^  +  2ab  +  b^. 
Hence f  the  second  power  of  the  sum  of  two  quantities,  contains 
the  second  power  of  the  first  quantity,  plus  twice  the  product  of 
the  first  by  the  second,  plus  the  second  power  of  the  second. 


16  MULTIPLICATION    OF    POLYNOMIALS.  ,XI 

Suppose  the  two  numbers  12  and  3;  then,  (12-|-3)  (12  +  3' 
=  144  -f-  72  -|-  9  =  225.  In  like  manner,  the  second  power  of 
3ab  +  2c  or  (3a6-f2c)  (3  ab -{-^c)  =  Qa^b^  +  12  ab^ 
+  4c2. 

Art.  3^    Find  the  second  power  of  a  —  h 
Operation, 
a  —  b 
a^b 
a^  —  ab 
—  ab  +  b^ 


This  differs  from  the  second  power  of  a  -|-  6  only  in  the  sign 
of  2  a  6,  twice  the  product  of  the  two  quantities,  which  is  in  this 
case  minus. 

Suppose  the  two  numbers  8  and  2 ;  then,  (8  —  2)  (8  —  2)  =r 
64  —  32  +  4  =  36.  In  the  same  manner,  (6bc  —  2m)  X 
(6  6  c  —  2  m)  =  3662  c2  —  24  6  c  m  +  4^2. 

Art.  36.  Let  the  learner  multiply  the  second  power  ol  a  +  ft 
by  a  +  6 ;  the  result,  that  is  the  third  power  of  a  +  6,  is 
a3  +  3a26  +  3a62  +  63, 

HencCf  the  third  power  of  the  sum  of  two  quantitieSy  contains 
the  third  power  of  the  first  quantity ^  plus  three  times  the  second 
power  of  the  first  into  the  second,  plus  three  times  the  first  into 
the  second  power  of  the  second,  plus  the  third  power  of  the  second. 

Art.  3T.  We  find,  in  like  manner,  that  the  third  power  of 
a  —  6  is 

«3_3a26  +  3a52_63^ 

which  differs  from  the  third  power  of  a  +  6  only  in  the  signs  o* 
he  second  and  fourth  terms,  which  in  this  case  are  negative. 


KU.  DIVISION   OF    IVIONOMIALS.  47 


SECIION   XII 

DIVISION    OF    MONOMIALS    OR   SIMPLE    QUANTITIES. 

Art.  38.  In  multiplication,  two  factors  are  given,  and  it  is  re- 
quired to  find  the  product ;  whereas,  in  division,  one  factor  and 
the  product,  that  is,  the  divisor  and  dividend,  are  given,  and  it  is 
required  to  find  the  other  factor  or  quotient. 

Division  then  is  the  reverse  of  multiplication ;  and  it  is  evi- 
dent, that  the  divisor  and  quotient  multiplied  together,  must  re- 
produce the  dividend.  For  this  purpose,  the  coefficient  of  the 
quotient  must  be  such,  as  multiplied  by  the  coefficient  of  the  di- 
visor, will  produce  that  of  the  dividend ;  and  the  exponent  of  any 
letter  in  the  quotient,  must  be  such,  as  added  to  its  exponent  in 
the  divisor,  will  produce  the  exponent  of  that  letter  in  the  divi- 
dend. 

Divide  a  6  by  a,  or  find  the  -  part  of  a  b. 

The  quotient  is  6,  because  the  product  of  a  and  b  13  ab.  Or 
if  a  6  be  divided  by  a,  the  quotient  is  6,  for  the  same  reason. 

Ans.  3  c. 

Ans.  4  be. 

Ans.  1. 

Ans.  a. 

Ans.  1. 

Ans.  ab  c. 

Ans.  Si/. 

Divide  84 a  6  c  x  by  12  c.     Ans.  7 abx. 
The  above  answers  are  correct,  because  in  each  case  the  quo- 
tient multiplied  by  the  divisor,  produces  the  dividend. 
Divide  a''  by  a*.     Ans.  a? ;  because  a*  .  a^=:^  a7. 
Divide  3 a^  b^  by  a b^     Ans.  3  a^  fts .  because  Sa^b^  ,ab'^  = 

Art.  30.  In  the  multiplication  of  monomials,  when  the  same 
letter  occurred  in  both  factors,  the  exponents  of  that  letter  were 
idded;  and  we  see,  from  the  last  two  examples,  that,  in  div  sion^ 


Divide  Sab  c 

by  ab. 

Divide  Sbcx 

by2x. 

Divide  a  or  1  a 

by  a. 

Divide  a  or  1  « 

by  1. 

Divide  ab 

by  ab. 

Divide  abc 

by  1. 

Divide  21  x  y 

by  7x. 

48  DIVISION    OF    MONOMIALS.  XII 

when  the  same  letter  occurs  in  both  dividend  and  divisor,  the 
exponent  of  that  letter  in  the  latter,  must  be  subtracted  from  its 
exponent  in  the  former. 

If  a  be  divided  by  a,  the  quotient  is  1 ;  for,  any  quantity  di- 
vided by  itself,  gives  unity  for  a  quotient.     But,  if  we  perform 

the  division  by  subtracting  the  exponents,  we  have  -    or  -r    = 

aK  Hence,  (ax.  7),  «°=  1.  That  is,  any  quantity  with  zeri 
for  an  exponent  is  equal  to  1. 

Art.  40.  From  the  preceding  examples  and  observations,  we 
derive  the  following 

RUIiE    FOR    DIVIDING    ONE   MONOMIAl.   BY    ANOTHER. 

1.  Divide  the  coefficient  of  the  dividend  by  the  coefficient  of 
the  divisor. 

2.  Strike  out  from  the  dividend  the  letters  common  to  it  and 
the  divisor,  when  they  have  the  same  exponents  in  both ;  but,  if 
the  exponents  of  any  letter  are  different,  subtract  its  exponent  in 
'Ac  divisor  from  that  in  the  dividend,  and  write  the  letter  in  the 
quotient  with  an  exponent  equal  to  the  remainder. 

3.  Write  also  in  the  quotient,  with  their  respective  exponents^ 
the  letters  of  the  dividend  not  found  in  the  divisor. 

Remark.  If  in  any  case,  however,  the  divisor  and  the  •divi'i 
dend  are  equal,  the  quotient  will  be  1. 


1. 

Divide  36  a^  b^  c^  m 

by  4  a3  fe  c^.     Ans.  9ab^m. 

2. 

Divide  4S«3  63  c2rf 

by  12  «  62  c. 

a 

Divide  150  a^b^cd^ 

by  30  a3  65^. 

4. 

Divide  120  a  6?  A3 

by  10  aft. 

5. 

Divide  125  m^  x  y'^ 

by  5  X  y. 

6. 

Divide  93  d^  b^  m^  x 

by  3  a  6  »i  x. 

7. 

Divide  lUm^a^Wxy^ 

by  37  wix  3^2  63. 

8. 

Divide  27  a2A2„a 

by  27. 

9. 

Divide  15  a  m^  x 

by  a  m^  x. 

10. 

Divide  27  a^m^x*y 

by  3  a  »|5  x. 

11. 

Divide  729  x<y4 

by  9x3  y. 

12 

Divide  IQa^b^x^ 

by8a62i» 

XIII.                                 DIVISION  OF    POLYNOMIALS.                                        49 

13.  Divide  99mx^f  z^  by  33  x  y*  z^. 

14.  Divide  1008 aP  h^  c^ d^  x  hySahc^ x. 

15.  Divide  111  »i5/i6p7  by  3/»^w5. 

16.  Divide  115 w^r^s^  by5nrs. 

17.  Divide  75x^0^13  by  25x8^3 
18  Divide  350  a?  614^7  by  50  a*  x^. 

19.  Divide  790  w^i  wi2  by  lOm^n?. 

20.  Divide  927  xi5y  hy^x^^y. 


SECTION   XIII. 


DIVISION    OF    POLTNOMIALS. 


Art.  41.  Iffl  +  6  —  cbe  multiplied  by  m,  the  product  ia 
am-\-hm  —  cm]  therefore,  \^  am-\-hm  —  cm  be  divided  by 
m,  the  quotient  is  a-^h  —  c. 

A  simple  quantity  is  a  factor  of  a  polynomial,  when  it  is  a  fac- 
tor of  every  term  of  that  polynomial ;  and  the  division  of  a  poly- 
nomial by  a  simple  quantity,  consists  merely  in  dividing  each 
term  of  the  former  by  the  latter.  But  a  question  arises,  how  we 
are  to  determine,  in  all  cases,  the  signs  of  the  partial  quotients. 
The  following  considerations  will  enable  us  to  decide  that  point. 

If  +  a  ^  be  divided  by  -|-  «>  the  quotient  will  be  -\-b,  because 
-j-  a  multiplied  by  -\-b  gives  -\-ab. 

If -|-a&  be  divided  by  — a,  the  quotient  will  be  — 6,  because 
—  a  multiplied  by  —  6  gives  -\-ab. 

If  —  ab  be  divided  by  -f-  a,  the  quotient  will  be  —  b^  because 
ya  multiplied  by  — b  gives  — ab. 

Lastly,  if  —  ab  be  livided  by  —  a,  the  quotient  will  be  -|- i, 
tiecause  —  a  multiplied  by  -|-  6  gives  —  ab. 

Hence,  the  rule  for  the  signs  in  division,  is  the  same  as  that  in 
multiplication  ;  that  is,  if  the  two  terms ,  one  of  which  is  to  be  di" 
mded  by  the  other y  have  the  same  sign,  either  both  -f-  or  both  — , 
the  quotient  must  have  the  sign  -j- ;  but  if  they  have  different 
signs f  that  is,  one  -\-  and  the  other  — ,  the  quotient  must  have  thi 
nrn  — . 

5 


50  DIVISION    OF    POLYNOMIALS.  XlII 

Art.  42.  Hence,  we  have  the  following 

RULE    FOR    DIVIDING    A    POLYNOMIAL    BY    A    MONOMi  IL. 

Divide  each  term  of  the  dividend  by  the  divisor ^  according  to 

hi   principles  given   in  Art.   40  for  dividing  one  monomial 

by  another,  observing  the  rule  established  above  for  the  signs ; 

and  the  partial  quotients  taken, together  ^  will  form  the  entiic  quo 

tient, 

1.  Divide  a^  b -]- a  m  by  a. 

2.  Divide  I2x  i/ —  6x^ -{-Sx  y^  by  3  x. 

3.  Divide  15  a  m^  +  30  m^  z  —  45  m^  by  15  m^. 

4.  Divide36  62ce/— 24  63 c2— 12  64c  by  662  c. 

5.  Divide  9  a5  66  _  3  ^2  63  by3«2  63. 

6.  Divide— 21x2 3^  —  7  + 42z  by— 7. 

7.  Divide  56  a^  63  c  —  28  a^  63—168  a?  6^  c^  by  —  28  a^  6-. 

Art.  43.  When  the  dividend  and  divisor  are  both  polynomi- 
als, the  process  becomes  rather  more  difficult ;  but,  if  we  observe 
(he  manner  in  which  a  product  is  formed  by  multiplication,  we 
shall  readily  see  the  course  to  be  pursued  in  division. 

Multiply  3  a3  +2a2  6  +  a 62  by  2a2  -|-  3a  6. 

Operation, 
3a3+   2a26  +  a62 
2a2-(-   3a6 


6a5+   4a46  +  2a362 

-f   9an+6a362-f  3q263 

6a5-j-13a4  6  +  8a3  62  4-3a2  63.   Product. 

If  this  product  be  divided  by  the  multiplicand,  the  quotient 
will  De  the  multiplier ;  or,  if  it  be  divided  by  the  multiplier,  the 
quotient  will  be  the  multiplicand. 

Since  each  term  in  the  multiplicand  is  multiplied  by  each  term 
in  the  multiplier,  if  no  reduction  takes  place,  the  number  of 
terms  in  the  product  will  be  equal  to  the  number  produced  b) 
multiplying  the  number  of  terms  in  the  two  factors  together 
Thus,  if  one  factor  have  4  terms  and  the  other  3,  the  product 
will  contain  12  terms.     In  most  cases,  however,   a   reduction 


^J.11.  DIVISION    OF    POLYNOMIALS.  51 

takes  place,  by  which  some  terms  are  united  and  others  wholly 
disappear. 

But  there  are  always  two  terms,  which  can  neither  be  united 
with  any  others,  nor  cancelled  by  any  others ;  viz :  one,  which 
is  produced  by  multiplying  the  term  containing  the  highest  power 
of  any  letter  in  the  multiplicand,  by  the  term  containing  the 
highest  power  of  the  same  letter  in  the  multiplier;  and  the  other, 
which  arises  from  the  product  of  the  terms  with  the  lowest  expo* 
uents  of  the  same  letter. 

Now,  since  the  dividend  is  to  be  considered  as  the  product  of 
the  divisor  and  quotient,  it  is  evident,  that  if  the  term  containing 
the  highest  power  of  a  particular  letter  in  the  dividend,  be 
divided  by  the  term  containing  the  highest  power  of  the  same 
letter  in  the  divisor,  the  result  will  be  the  term  of  the  quotient 
containing  the  highest  power  of  that  letter. 

J.et  us  reverse  the  process  of  multiplication,  and  divide  Ga^-j- 

Operation. 
Dividend. 

6q5-4-13a^6  +  8g3  62-|-3a2  63f  3a3  +  2a26  +  q52.  Divisor. 
6a5-[-   4a^6-f^«^^^ X'^a^  +  ^ab,    Quotient.. 

9a46+^^^  +  3a2  63 

9g^6-f  6«^^^  +  3flQ63 

0. 

According  to  the  preceding  remarks,  6  a^  divided  by  3  o^ 
must  produce  the  term  containing  the  highest  power  of  a  in  the 
quotient ;  3  a^  is  contained  2  a^  times  in  6  a^ ;  then,  as  the  en- 
tire dividend  is  produced  by  multiplying  the  whole  of  the  divisor 
by  the  whole  of  the  quotient,  if  we  multiply  the  whole  divisor  by 
this  first  term  2  a^  of  the  quotient,  a  part  of  the  dividend  will  be 
produced.  The  product  of  the  divisor  by  2  a^  is  QaP  -\-4,af^h 
-|-  2  a3  ft2j  which  being  subtracted  from  the  dividend,  leaves  ^a^h 
+  6a362^3a253. 

This  remainder  is  to  be  considered  as  a  new  dividend,  and  as 
produced  by  multiplying  the  divisor  by  the  remaining  part  of  the 
ijuotient.     The  first  term,   9  a*  6,  of  this  new  dividend,  must 


62  DIVISION    OF    POLYNOMIALS.  Xlli 

have  been  produced  by  multiplying  3  a^  by  the  term  containing 
the  next  highest  power  of  a  in  the  quotient ;  therefore,  if  it  be 
divided  by  3  a^y  that  term  of  the  quotient  will  be  obtained.  Di- 
viding 9 a'* 6  by  3a^,  we  have  3a6  for  the  second  term  of  the 
quotient ;  then,  multiplying  the  divisor  by  3  a  6,  the  product  is 
9  a"*  6  +  6  a3  6^  -|-  3  a^  ^3^  which  subtracted  from  the  last  divi- 
dend, leaves  no  remainder.  The  entire  quotient  therefore  is  2  a^ 
+  3a6. 

We  perceive,  from  the  preceding  example,  that  we  always  divide 
the  term  containing  the  highest  power  of  some  letter  in  the  divi- 
dend, by  the  term  containing  the  highest  power  of  the  same  letr 
ter  in  the  divisor  It  will,  therefore,  be  found  convenient  to 
write  the  quantities  in  such  a  manner,  that  these  two  terms  may 
stand,  the  one  on  the  left  of  the  dividend,  and  the  other  on  the 
left  of  the  divisor.  This  will  be  accomplished  by  arranging  the 
dividend  and  divisor  according  to  the  powers  of  the  same  letter, 
beginning  with  the  highest. 

A  polynomial  is  said  to  be  arranged  according  to  the  powers 
of  a  particular  letter,  when  the  terms  are  so  written,  that  the 
powers  of  that  letter  go  on  increasing  or  diminishing  from  left  to 
right.  Thus,  in  the  example  just  performed,  the  quantities  were 
arranged  according  to  the  diminishing  powers  of  a. 

With  regard  to  the  signs  of  the  partial  quotients,  the  same  rule 
Is  applicable,  that  was  given  for  the  division  of  a  polynomial  by 
a  monomial. 

Art.  4L4.   From  what  precedes  we  deduce  the  following 

RULE    FOR   THE    DIVISION    OF   ONE    POLYNOMIAL    BY    ANOTHER. 

1.  Arrange  the  dividend  and  divisor  according  to  the  powcs 
of  the  same  letter ^  beginning  with  the  highest. 

2.  Divide  thejirst  term  of  the  dividend  by  the  first  term  of  the 
divisor  y  and  place  the  result  as  the  first  term  of  the  quotient;  rec- 
ollectingy  that  if  both  terms  have  the  same  sign,  the  partial  quo- 
tient must  have  the  sign  -|-,  but  if  they  have  different  signs,  thi 
partial  quotient  must  have  the  sign  — . 

3.  Multiply  the  entire  divisor  by  this  term  of  the  quotient,  sub- 


XIII.  DIVISION    OP    POLYNOMIALS.  53 

tract  the  product  from  the  dividend,  and  the  remainder  will  form 
a  new  dividend. 

4.  Divide  the  first  term  of  the  new  dividend  hy  the  first  tern- 
of  the  divisor  J  and  the  result  will  form  the  second  term  of  the  quo^ 
tient ;  multiply  the  entire  divisor  hy  this  second  term  of  the  quo- 
tientj  and  subtract  the  product  from  the  second  dividend.  The 
remainder  will  form  a  new  dividend,  from  which  another  term  of 
the  quotient  can  he  found. 

These  operations  are  to  he  repeated,  until  all  the  terms  of  the 
>*rginal  dividend  are  exhausted. 

N.  B.  The  same  arrangement  must  be  preserved,  in  each  of 
the  partial  dividends,  as  was  made  at  first  in  the  whole  dividend. 

When  the  first  term  of  any  remainder  cannot  be  divided  by 
the  first  term  of  the  divisor,  the  process  must  terminate,  unless 
the  quotient  be  continued  in  a  fractional  form.  When  the  divi- 
sion terminates,  the  remainder,  if  there  be  one,  may  be  written 
over  the  divisor,  in  the  form  of  a  fraction,  and  annexed  to  the  en- 
tire part  of  the  quotient. 

Let  it  be  proposed  to  divide  75a^h*  —  27  a  h^  —  49a^b^-\- 
20a^h  —  l9a*h^hy  4a^b  +  2h^^7ahK 

Operation. 
20a56_i9a462__49a353_^75a254__27a65  (  4.a^h^7ah^+Sb^ 

20a56_35a462^15a353 \  5a^+^a%—9ah^ 

+ 1 6a462— 64fl363_^75a264— 27a65 

-fl6q^6Q— 28a3&3+12a^6^ 

_36a363^03a264_27a65 
_36«363_|_63a254_27g55 

0. 
After  having  arranged  the  two  quantities  according  tc  the 
powers  of  a,  and  placed  the  divisor  on  the  right  of  the  dividend, 
we  divide  20  a^  6  by  4  a^  6,  which  gives  -\-5a^  for  the  first  term 
of  the  quotient ;  we  then  multiply  the  divisor  by  5  a^,  write  the 
product  under  the  dividend,  and  subtract  it  from  the  dividend. 
The  subtraction  of  the  product  is  performed  by  changing  itj 
signs,  considering  it  as  written  after  the  dividend,  and  redu- 
cing similar  terms ;  thus,  the  signs  being  changed,  it  becomes 
5* 


54  DIVISION    OF    POLYNOMIALS.  XIII. 

—  *20a5  6_|_35a4  52_i5a3  53.  then,  by  reduction,  +20a5j 
and  — 20a^b  cancel  each  other,  — 19  a^  6^  and  -{-25  a*  b* 
make  +  16a4  62,  and  —i9a^b^  and  —  15  a^b^  make  — 64^3 fts. 
bringing  down  the  other  terms  of  the  given  dividend,  we  have  for 
a  remainder  IGo^i^ — 64 a^ 63 _j_ 75 ^2 M  —  27^65^  which  forms 
a  new  dividend 

We  now  divide  16  a^b'^  by  the  first  term  of  the  divisor  4  a^  6, 
and  obtain  for  the  second  term  of  the  quotient  +  4  a^  6  ;  multi- 
plying the  divisor  by  4  a^b^  and  subtracting  the  product,  -J- 16  a^  6* 

—  28  a3  ^3  -[-  12  a2  ^4^  from  the  second  dividend,  in  the  same 
manner  as  before,  we  obtain  for  a  remainder  — 36  a^b^-\-63a^b^ 

—  27  a  6^,  which  forms  the  third  dividend. 

We  now  divide  —  36  a^  b^  by  4  a^  6  and  obtain  —  9ab^  for 
the  third  term'  of  the  quotient ;  multiplying  the  divisor  by  — 
9  a  62,  and  subtracting  the  product,  —  36  a^  63  -[-  63  a^  6^  — 
27  a  b^y  from  the  third  dividend,  we  have  no  remainder.  The 
entire  quotient,  therefore,  is  5  a3  -|-  4  a^  6  —  9  a  62. 

As  another  example,  let  it  be  proposed  to  divide  a^  —  6^  by 
a  — 6. 

Operation. 


a5  — 

a^b  i  a^-\-a^b  +  a^b^  +  ab^ 

+  6- 

+ 

aH  —  b^ 

+ 

a^b  —  a^b^ 

+  a3  62_55 

4-a3  62_a2  53 

+  «2  63-65 

-|-a2  63_a64 

+  a64  — 

65 

+  a  64  — 

65 

0. 

Ill  the  last  example,  several  terms  are  produced  in  the  course 
of  the  operation,  which  are  not  found  in  the  dividend ;  these 
terms  disappear,  by  reduction,  when  the  quotient  and  divisor  are 
multiplied  together. 

1.  Divide  a3  -|-  3  a^  m  +  3  a »»'  + 1»3  by  a  +  m. 


ZCITI.  DIVISION    OF    POLYNOMIALS.  55 

2.  Dividea3  +  5a4x+10a3  22_[_i0a2x3  +  5ax4-f.x5  by 

3.  Divide  m"^  —  4  m^  z  -|-  6  w^  x^  —  itnx^  -^-x^  by  m  —  x. 

4.  Divide  (3 x'*  — 96  by  3z  + 6. 

5.  Divide264  — 9a2i2^6a4  +  4a36  — 3a63  by  2a^  + 

a  Divide20a5_4ia4  5_|_50a3  62__45fl2  53_|.25a64  — 
665  by  5a^  —  Aa^b  +  5ab^  —  Sb^ 

7.  Divide4x4  —  9a22;2+6a3x  — a4  by  2x2  +  3izx  — a^ 

8.  Divide  a^ -{- 2  a^  z^ -^- z^  by  a^  —  a  2  +  22. 

9.  Divide  a^  __  I6  a3  2;3  _|_  64  x^  by  a2  _  4  ^  x  +  4  x2. 

10.  Divide  x^  —  x^  +  x3  —  x2  +  2  x  —  I  by  x2  -(-  x  —  1. 

11.  Divide  10a4  _  48  a3  6  _^  51  ^2  62  +  4  a  63  —  15  6^   by 
8a6  — 2a2_565J. 

12.  Divide  ni^  —  x"^  by  we  —  x. 

13.  Divide  5a5  63c5__22a4  63  c6_^5a363c7^12a263c8_ 

'7«262c8_|_28a62c9  by  a26c2_4a6c3. 

Art.  45.    In  the  preceding  examples,  the  division  could  be 
exactly  performed.      Let   it   now  be  proposed  to  divide  a   by 

Operation. 
l—x 


a  /  1  — X 

a  —  ax  (  a-\-a 

+  ax 

ax  —  ax^ 


X  +  «  t;^  +  a  x3  -(- 

1  —  X 


+  ax2 

a  x2  —  a  x3 
+  ax3 

a  x3  —  ax* 

+  a  X*.. 

i  «  the  exanjple  above,  it  is  manifest  that  the  operation  would 

never  terminate.     This  quotient  is  called  an  infinite  series,  and 

we  see  that  any  term,  except  the  first,  may  be  formed  by  multi' 

plying  the  preceding  term  by  x,  remembering  to  place  the  divi' 


56  DIVISION    OF    POLYNOMIALS.  XllJ 

sor  under  the  last  remainder,  to  indicate  the  continuation  of  th<! 
division.     The  mode  in  which  any  term  may  be  found  by  means 
of  the  preceding  term,  is  called  the  law  of  the  series. 
Let  us,  for  a  second  example,  divide  1  by  1  -[-  «. 
Operation. 


l  +  a(l  — a  +  aS 


Jf.a*  —  ai-ff.    "* 


1  +  a- 
a 


a^ 

a^  + 

a3 



a3 

a3_ 

-a* 

-— 

a5_ 

-(fi 

C6. 

Here  we  perceive,  that  the  law  of  the  series  is  the  same  as.  be 
fore,  except  that  the  2d,  4th,  6th,  &,c.  terms  have  the  sign  — . 

Art.  4:0.  The  difference  between  similar  powers  of  two  quan* 
tities,  the  exponent  of  the  powers  being  integral  and  positive ^  is 
divisible  by  the  difference  of  those  quantities. 

Thus,  a"*  —  6*"  is  divisible  by  a  —  b. 

To  prove  this,  it  is  only  necessary  to  find  the  first  remainder 

Operation. 
a«  —  5»»         la  —  6 


First  remainder,  a**"'^  —  6*"=:6(a'*-i  —  6'*"^). 

Now  it  is  manifest,  that  if  this  remainder,  6(a'"~i  —  6"*-!) 
tan  be  exactly  divided  by  a — ft,  the  dividend,  a*"  —  6"*,  is  divisi 
Ue  b/  it  also.     But  since  a  product  is  divided  by  dividing  one 


XIIL  DIVISION    OF    POLYNOMIALS.  51 

of  its  factors,  this  remainder,  and  consequently  a"*  —  6"*,  is  di 
visible  by  a  —  b,  if  the  factor  a"*"^  —  b'"~^  is  divisible  by  it. 

That  is,  if  a  —  b  divide  cf"^  —  6"'~i,  it  will  also  divide  a'"  — 
6"*;  or,  in  other  words,  if  the  proposition  is  true  for  the  m  —  1th 
power,  it  must  be  true  for  the  mth  or  next  higher  power. 

But  we  have  already  seen  (Art.  44),  that  it  is  true  for  the  5th 
power  ;  therefore,  it  is  true  for  the  6th ;  being  true  for  the  6th, 
it  must  also  be  true  for  the  7th,  and  so  on.  Hence,  it  must  be 
true  in  all  cases,  and  a^  —  6"*  is  divisible  by  a  —  b. 

In  like  manner,  (a  +6)"*  —  (c  +  rf)*"  is  divisible  by  (a-\-b) 
-(c  +  d). 

We  continue  the  operation  of  dividing  a*"  —  ft*"  by  a-^b,  in 
order  that  the  learner  may  see  the  form  of  the  quotient. 
oT  —  6""  t  a  —  b 

i  a'-- 1  +  a—2  b  +  ar-^  b^ a  b^-^  +  i""! 

ar-ib  —  b"* 


oT-^b^  —  6^ 

Of  course,  the  number  of  terms  in  the  quotient  must  be  in- 
definite, until  some  determinate  value  is  assigned  to  m.  The 
points are  used  to  supply  the  place  of  the  indefinite  num- 
ber of  terms.  It  will  be  easy  to  perceive  from  what  follows,  that 
tie  last  two  terms  must  be  such  as  we  have  represented  them. 

It  may  be  proved,  however,  that,  in  any  case  like  the  prece- 
ding, the  number  of  terms  in  the  quotient,  will  always  be  equal 
to  the  exponent  m.  This  fact  may  be  deduced  from  an  examina- 
tion of  the  successive  terms  of  the  quotient. 

Since  the  division  can  be  exactly  performed,  the  last  term  of 
the  quotient  must  be  such,  as,  being  multiplied  by  6,  the  last 
term  of  the  divisor,  will  produce  6"*,  the  last  term  of  the  divi- 
dend; that  is,  it  must  be  6*""^,  for  6  times  ft'""^  is  6**. 

But  we  see,  that,  in  the  successive  terms  of  the  quotient,  the 
exponent  of  a  goes  on  diminishing  by  unity  as  many  units  being 


58  DIVISION    OF    POLYNOMIALS  X.  11 

subtracted  from  m  in  each  case,  as  mark  the  number  of  the  term 
from  the  first  inclusive ;  and,  that  the  exponent  of  h  in  any  case, 
is  always  1  less  than  the  number  of  the  term. 

Consequently,  in  the  mth  term,  the  exponent  of  a  must  be 
m  —  mot  0,  and  the  exponent  of  h  must  be  m  —  1.  The  7/ith 
term,  therefore,  is  a^b^-^  or  ft*""!,  (Art.  39).  But  we  have 
seen  that  the  last  term  must  be  6*""^.  Hence  the  mth  term  and 
the  last  term  are  the  same;  in  other  words,  there  are  m  terms  in 
the  quotient. 

Thus,  i-ZLL  =  x^  +  x^ y  +  x^ y^  +  z^ y^  -\-x^ y^  +  z^ y^  + 
X      y 

t  y®  -}-  y^ ;  the  quotient  containing  8  terms. 

It  might  also  be  proved,  that  the  difference  between  similar 

even  powers  of  two  quantities,  and  the  siim  of  similar  odd  powers, 

are  each  divisible  by  the  sum  of  those  quantities. 

Thus,  —  gives  an  exact  quotient  of  n  terms,  when  n  is 

an  even  number ;  and  — J~  gives  an  exact  quotient  of  n  terms, 
i^he.i  n  is  an  odd  number. 

1.  Divide  a;9  —  y^  by  x — y. 

2.  Divide  243  a5  — 1024  65   by  3  a  — 4  6. 

3.  Divide  xio  — yio  by  x  +y. 

4.  Divide  1  —  m^  by  1  +  m. 

5.  Divide  m^  -\-n^  by  m  -|-  w. 

6.  Divide  m^  -\-\  by  m+  1. 

7.  Divide  I  by  1  —  m^,  finding  6  terms  of  the  series,  and  r  > 
nexing  the  remainder  placed  over  the  divisor. 

8.  Divide  a  by  \-\-xy^  finding  6  terms  of  the  series,  and  an 
nexing  the  remainder  placed  over  the  divisor. 

Art.  -47.  It  is  manifest,  that  when  a  product  is  represented 
in  its  factors,  dividing  one  of  the  factors,  divides  the  whole  pro- 
duct; also,  that  we  may,  in  any  case,  divide  the  dividend  first  by 
one  factor  of  the  divisor,  then  divide  the  resulting  quotient  b> 
another,  and  so  on. 


XIV  MULTIPLICATION    OF    FRACTIONS.  69 

Thus,  7  8  .  3  =  168 ;  dividing  the  factor  8  by  4,  we  have 
7  .  2  .  3  zn  42,  which  is  ^  of  168. 

Also,  in  dividing  6  .  12  or  72  by  2  .  4  or  8,  we  may  first  di 
vide  by  2,  which  gives  3 .  12,  and  then  this  quotient  by  4,  whicb 
gives  3.3  =  9  =  ^. 

In  like  manner,  (m  —  n)  (a^  —  6^)  divided  by  a  —  6,  gives 
{m  —  n)  {a  -\-  6),  to  obtain  which,  we  divide  the  factor  a^  —  b^ 
by  the  divisor. 

Also,  (x2  _  y2)  (a2  _j_  2  a  6  _|_  ^2)  divided  by  (x  +  y )  (a  +  6), 
gives  (x — y)  (a-^-b);  to  obtain  which,  we  divide  the  factor 
a;2  —  y2  by  2;  _j_  y  ^  and  the  factor  a^  _|_  3  «  5  _|_  52  ^y  a  +  6. 

1.  Divide  a^C^J  +  y)  ^Y  ^ -{-!/• 

2.  Divide  (m2  — l)(rt  +  6)  by  m  — 1. 

^.  Divide  14(a;3_l)  (a  +  ni)  by  7(a;  — 1). 

4.  Divide  («2_62)  (a:3_|.y3)  by  (a  — 6)  (x  +  y). 

^.  Divide  27  (m^  —  n^)  (x^  +  yS)  by  3  (m^  +  n2)  (i  +  y ). 

6.  Divide  30  m2  (x^  —  yS)  (;;j4__„4) 

by  10w(x3  +  y3)  (m2  — w2). 
7    Divide25(a  +  6)  (m4— 1)  (x5+l) 

by5(a  +  6)(m2+l)(x+l). 


SECTION    XIV. 

MULTIPLICATION    OF    FRACTIONS    BY    INTEGRAL    QUANTITIES. 

Art.  4:8.    Fractions  have  the  same  signification  in  algebra, 

that  they  have  in  arithmetic.     Thus,  -  signifies  that  one  unit  is 

divided  into  b  equal  parts,  and  that  a  of  those  parts  are  useil ; 
or,  it  represents  division,  and  signifies  that  a  is  divided  into  i 
equal  parts. 

How  much  IS  5  times  -^j  ?     Ans.  -f ^. 

How  much  is  3  times  y  «     Ans.  — -. 
6  6 


-80  MULTIPLICATION    OP    FRACTIONS.  XIV 


How  much  is  c  times  -?     Ans.  -r-. 
0  o 

What  is  I  of  4?    |  of  4  is  ^,  and  f  of  4  is  ^.  Ans. 

What  is  ^  of  a?    ^  of  a  is  -,  and  f  of  a  is  — -.  Ans, 

TTTi      '     1     ^  /.»l/..c         .  a    „    .    ac 

What  is  the  -r  part  of  c  f     -  of  c  is  — ,  and  -  of  c  is  -^. 
o  6  0  0  o 

In  the  first  three  questions,  the  object  was  to  multiply  a  frao 
tion  by  an  integral  quantity,  and  in  the  last  three,  to  find  a  frac- 
tional part  of  a  quantity,  that  is,  to  multiply  an  integral  quantity 
by  a  fraction ;  and  we  perceive  that  both  objects  were  accom- 
plished by  multiplying  the  numerator  and  the  integral  quantity 
together.  •* 

Hence,  to  multiply  a  fraction  hy  an  integral  quantity ^  or  an 
integral  quantity  hy  a  fraction  ;  multiply  the  numerator  hy  thi 
integral  quantity,  and  write  the  product  over  the  denominator, 

1.  Multiply  —  by  6  c, 

2.  Multiply  by  mx, 

3.  Multiply  •  by  m-\-n, 

4.  Multiply  -^^  by  4c  +  3a;. 

5.  Multiply  by  wi-j-n. 

6.  Multip.y  r—-^ by  12a3  +  25a2  6. 

^    h^-\-^ac 

7.  What  is  the  ^-i^  pa^  of  a^--h^1 

h  -\-  c 

a   What  is  the     o^To"^^^''  o  part  of  2  a  c  -|-2 ft  cl 
9    Multiply  25  x2  +  13  z  y  by  '^^^~^-. 


X.IV  MULTIPLICATION    OF    FRACTIONS.  61 


10.   Multiply  ^  by  5. 
15 

The  fraction  —  signifies  that  a  is  divided  into  15  equal  parts . 

but  if  the  denominator  be  divided  by  5,  a  will  then  be  divided 

into  3  parts,  or  ^  as  many  parts  as  it  was  before  ;  consequently, 

each  of  the  parts  will  be  5  times  as  great  as  before ;  that  is,  the 

a  .    ^    .  a 

fraction  ^\ao  times  --. 
o  15 


1 1.   Multiply  r-  by  h. 


a 
c 
Here  a  is  divided  into  6  c  equal  parts ;  but  if  the  denominator 

be  divided  by  6,  a  will  then  be  divided  into  t  as  many  parts  as  it 

was  before ;  the  parts,  therefore,  will  be  h  times  as  great  as  they 

were  before:  that  is,  the  fraction  -  is  6  times t-. 

c  be 

HevCf  to  multiply  a  fraction  and  an  integral  quantity  togetJir 
er,  divide  the  denominator  by  the  integral  quantity  ^  if  possible. 

Art  49.   Combining  this  rule  with  the  preceding,  we  have  a 

GEN-J  AL     RJIiE     TO     MULTIPLY    A    FRACTION    AND    AN     INTE6Rii.l- 
QUANTITY    TOGETHER. 

Divide  the  denominator  by  the  integral  quantity ,  if  it  can  be 
dona ;  if  noty  multiply  the  numerator  by  the  integral  quantity. 

The  following  examples  may  be  performed  by  dividing  the 
denominators. 

t    Multiply  — 5^  by  m. 

3  a  5  4-  4  a  V 

2.  Multiply  z^  y  by  ^g  ,3  y.  1 19  ^4  y»- 

3.  Multiply  !!!!=^±^*  by  a  +  6. 

4    Multiply Xi^^  +  25  6°  +  43 

6 


MULTIPLICATION    OF    FRACTIONS.  XIV 


6-   Multiply  ^j-p^^^i^l+J^^j^^,-^,  by  a3  +  3«M 
7.   Multiply  ^^l^j  by  .=  +  ,=». 

9-  Multiply  ^^^,"^^^3^  by  m  +  y. 

10-  Multiply  ^^-^^.i^^ 

^1-  Multiply  ,^  (,!+)\:r,,.)  by  7(a-6)  (c^d), 

^^'  Multiply  (^.!.'J)|jj^3^5)  by  (m^n)  (x  +  y). 

13.  Multiply  T  by  6. 

Dividing  the  denominator  by  6,  the  fraction  becomes  -  or  a 

14.  Multiply ^ by  x^  y. 

Dividing  the  denominator    by  x^  y,   the  fraction   become^' 
!!^±A£,thatis,m2  +  6c. 

From  the  last  two  examples  we  see,  that. 

If  a  fraction  is  multiplied  by  a  quantity  equal  to  its  denomh 
fMtor,  the  product  will  be  its  numerator, 

15.  Multiply  ^^^  by  c. 

16.  Multiply  — ^  _A         by  a  — 6. 


17.   Multiply  3^^^^  +  ^=^y  by  m^  +  Sme. 
^^      m^-\-3mc       ^         ' 


XIV  MULTIPLICATION    OF    FRACTIONS.  f>3 

18.   Multiply  — ^  by  127  a. 

(Z  c 

Art  50.    Multiply  — g—  by  mi/. 

First  multiply  by  m ;  the  product  is  ;   multiply  this  pro- 

mx 

duct  by  y,  and  the  result  is  — -  (Art.  40). 

The  result  would  have  been  the  same,  if  we  had  divided  the 
integral  quantity  and  the  denominator  of  the  fraction  both  by  m, 
and  then  proceeded  according  to  the  first  rule  in  Art.  4:8. 

Therefore, 

When  an  integral  quantity  and  a  fraction  are  to  he  multiplied 
together,  if  the  integral  quantity  and  denominator  of  the  fraction 
have  common  factors,  those  factors  may  he  omitted  in  both  before 
multiplying. 


1. 

Multiply  ^2^2 

hymxy. 

2. 

Multiply  j2 ,3  rf 

hy  ah^c\ 

3. 

Multiply  ac^xy 

<c-     . 

4. 

Multiply  7  a3x3 

"S? 

5. 

Multiply  J^ 

by  3nfix. 

6 

Multiply  3  7715^2  (05  _f_  5) 

a^x3 

^6m3(a  +  4)- 

r. 

Multiply  .,_  .  i.x?'      „ 

-,hj  i(a  +  b)lc-d). 

8-   Multiply  jg^^3:^j  by7ac(i  +  y;. 

9    Multiply  3(.+m)(<»+.)  by  ^ (,.  +  gtl  + '»T 


64  DIVISION    OF    FRACTIONS.  XV 


SECTION   XV. 


DIVISION   OF    FRACTIONS    BY    INTEGRAL    QUANTITIES. 

Art.  51.   Divide  y\  by  2 ;  or  find  ^  of  ^.     Ans.  ^\ 

Divide  -j-  by  3 ;  or  find  X  of  -7-.     Ans.  — 
o  0  o 

■rv-i<356,     _  -.l.afe        .         a 

Divide  —  by  o ;  or  find  7-  of  — .     Ans.  -. 
c      "^  be  c 

For,  in  each  of  these  examples,  if  the  quotient  be  multiplied 
by  the  divisor,  the  product  will  be  the  dividend. 

Hence t  to  divide  a  Jr action  hy  an  integral  quantity ^  divide  the 
numerator  by  the  integral  quantity  ^  if  possible. 

1.  Divide  — —  by  2  a  c 

21  x2 1/3 

2.  Divide    '   ^  by7x2y. 

3.  Divide  -j— ^ f  by  60  a^  x. 

Am^  —  n" 

^    ^.  .^       17a2  59 

4.  Divide  7 r-: by  a^b\ 

o  c  4-4  zy 

5.  Divide  — ' by  a. 

m 

^•^^^^^^ — Ti6+cT —   '^^^• 

7.  Divide -^+"1^  +  ^^  bya+t. 

aji 53 

8.  Divide  tt; by  a — 6. 

36c  —  xy 

^   T..  .J    12x34-29z2  +  14x  u    A     I'f 

9   Divide  ^^^^^  by  4X  +  7. 


10 


^.  .^    18x3  — 33x2  +  44x  — 35,  o^ij. 

Divide  =^i j-^-r-s by  3x»  — 2i  +  5 

llm  +  4n2 


11    Divide  ^  by  c. 

o 


XV  DIVISION    OF    FRACTIONS.  65 

In  the  last  example,  we  cannot  divide  the  numerator ,  but  in 
Art.  48,  we  showed  that  a  fraction  is  multiplied  by  dividing  its 
denominator ;  on  the  other  hand,  a  fraction  may  be  divided  b} 
multiplying  its  denominator. 

The  fraction  —  signifies  that  a  is  divided  into  b  equal  parts, 

and  if  the  denominator  be  multiplied  by  3,  for  example,  a  would 
then  be  divided  into  3  times  as  many  parts,  and  the  parts  would 

be  only  ^  as  great  as  before ;  that  is,  5-=-  is  ^  of  rr.    In  like  man- 

S  b  o 

net,  if  the  denominator  be  multiplied  by  c,  a  will  be  divided  into 

c  times  as  many  parts,  and  the  parts  will  be  only  -   as   great   as 

c 

before  ;  that  is,  7 —  is  -  of — . 
be       c       b 

Hence,  to  divide  a  fraction  by  an  integral  quaidity,  multiply 
the  denominator  by  the  integral  quantity. 

Art.  5!3.  Combining  this  rule  with  the  preceding,  we  have 
the  following 

OSnVERAL     RUliE     FOR     DIVIDING     A     FRACTIOW     BY    AN     IITTEORAI. 
QUANTITY. 

Divide  the  numerator ,  if  it  can  be  done ;  if  not,  multiply  tht 
denominator f  by  the  integral  quantity, 

1.  Divide  7—  by  m. 

be  ' 

2l  Divide  ^i^  by6  +  c. 

b —  c  •' 

3.  Divide  ^i^  by  4x2. 

3a6  +  46 

4.  Divide  -rr — —  by  3  2;  +  4  m 

3z  —  4w 

6    Divide  — - — /    '  ^    by  ar-f-y- 

rf'H,  ,y2 

6.  Dvde        .  \  byx— y. 

a-\-b  '  ' 


6 


86  DIVISION    OP   FRACTIONS.  XV 

7  Divide  .  by  p  4-  or 

4  -\-.b  c 

8  Divide  ~^-. j — -    by  X  —  y. 

^-  ^^^*^«    62.{:c2      by  7. 

10.  Divide g —  .  by  x  —  y. 

11,  Divide ; by  w+?t. 

x  +  y  ^ 

Art.  53.   Divide by  6  w  w. 

7xy      ^ 

The  divisor  is  the  same  as  2  .Smn.  First  divide  by  2  w ; 
the  quotient  is .     Divide  this  quotient  by  3n,  the  remain 

5a9 
inff  factors  of  the  divisor;  the  result  is  — . 

The  result  would  have  been  the  same,  if  we  had  first  divided 
the  numerator  of  the  fraction  and  the  integral  quantity  both  by 
2  m,  their  common  factors,  and  then  proceeded  according  to  the 
second  rule  in  Art.  51. 

HencCf  when  a  fraction  is  to  be  divided  by  an  integral  quan^ 
tityy  if  the  numerator  of  the  fraction  and  the  integral  quantity 
have  common  factors  f  those  factors  may  be  omitted  in  both,  pre* 
vious  to  the  division. 

1.  Divide T —  by  3 ox. 

2    Divide  ^i^^^  by5a6x. 

27  »|3  x2  w3 

3.  Divide     Vf  ^J       hy  9 abmxy^. 

36x2 

4.  Divide— -5-  by36ax« 

7y2 

14  m3n 

5.  Diviffs ■ m^trnx 


SIVI  PRIME    FACTORS.  flT' 


6.  Divide  '^^('^  +  ^)  by  4  m  (a 4-  6). 

7.  Divide     ^    ^  by  7»i(a;  +  y). 
8.Oividei(^^=0^+^by2a(.  +  ,). 

9    Divide  5lf±^I^!!z:!!l)  by  Szy(a  +  b)(m^n), 


SECTION    XVI. 

FACTORS    AND   DIVISORS   OF   ALGEBRAIC    QUAWTITIBS. 

•  Art.  5  J:.  A  prime  quantity  in  algetira,  like  a  prime  number 
m  arithmetic,  is  one  which  is  divisible  by  no  entire  and  rational 
quantity,  except  itself .  and  unity.  Thus,  «,6  and  a-\-h  aie 
prime  quantities;  but  ah  \s  not  prime,  because  it  is  divisible 
both  by  a  and  h. 

Two  quantities  are  said  to  be  prime  with  regard  to  each  other^ 
when  the  same  quantity  will  not  exactly  divide  them  both,  that 
is,  without  a  remainder.  Thus,  a  h  and  c  m,  although  neither  of 
them  is  a  prime  quantity,  are  prime  with  respect  to  each  other. 

It  is  to  be  observed,  however,  that  when  we  call  a,  6,  c,  &c., 
prime  quantities,  we  mean  simply  that  they  cannot  be  algebra- 
ically divided  by  other  quantities,  except  by  representation ;  but 
they  are,  strictly  speaking,  prime  quantities,  only  when  they  rep- 
resent prime  numbers. 

Sometimes  it  is  requisite  to  separate  quantities  into  their  prime 
factors.  This  operation,  in  monomials,  is  attended  with  no  diffi- 
culty; for  we  have  only  to  ascertain,  according  to  the  method 
usually  given  in  arithmetic,  the  prime" factors  of  the  coefficient, 
and  torepresent  them  as  multiplied  together  and  followed  by  the 
several  literal  quantities,  each  written  as  many  times  as  it  is  a 
\ctor 

For  example,  12  a^js  --  3  .  32  ^9  53  --  3  . 2   2  a  a  6  6  6.     Ic 


f)8  PRIME    FACTORS.  XVI 

this  rjuantity  the  dVTerent  prime  factors  are  3,  2,  a  and  6 ;  3  is 
contained  once,  2  twice,  a  twice  and  6  three  times  as  a  factor. 

The  prime  factors  would,  however,  be  sufficiently  indicated 
merely  by  ascertaining  those  of  the  coefficient ;  for  the  exponents 
show  how  many  times  the  letters  are  contained.iis  factors.  Thus, 
54a2  65c4  — 2.  3.3.3  «2  65  c4:=  2.33  a^Sc^. 

When  a  quantity  is  the  product  of  a  monomial  and  a  prime 
polynomial,  in  order  to  separate  it  into  factors,  it  is  only  neces- 
sary to  divide  it  by  the  greatest  monomial,  that  will  exactly  di- 
vide all  the  terms,  and  to  place  the  divisor,  separated  into  prime 
factors,  before  the  quotient,  the  latter  being  included  in  a  paren- 
thesis. 

Thus,  ah-^h'^z=ih(a-\-h)\  in  which  the  factors  are  b  and 
a  +  6.  In  like  manner,  46c2  +  8  62c  =  46c(c  +  2  5)  = 
2^  6  c  (c  -|-  2  6) ;  in  this  case,  the  prime  factors  are,  2,  6,  c  and 
c  +  26 

Let  tne  learner  separate  the  following  quantities  into  prime 
factors. 

1.  am  —  hm.     Ans.  m(a  —  h) 

2.  ah-^-ac — 2  am. 

3.  4ax-\-2xi/-{-l2x. 

4.  25x^-\-20x^y  —  15x^m. 

5.  8labc  +  27a^b^c-\-5ia^bc. 

6.  99m^xy+ 108 m2pgr-f  18 m^r. 

7.  l2ab-\-2iabc  —  SQabx. 

Art.  55.  When  a  quantity  is  the  product  of  two  or  morr  pol- 
ynomials, the  process  of  finding  its  factors  becomes  more  diffi* 
cult ;  but  there  are  cases  in  which  some  of  the  factors,  at  least, 
may  be  easily  ascertained. 

1.  Any  quantity  which  is  known  to  be  B.power  of  a  polynomial, 
may  be  separated  into  as  many  factors,  each  equal  to  that  polyno- 
mial, as  there  are  units  in  the  exponent  of  the  power.  Thus 
a^  +  2ab  +  b^z=z(a  +  b)^  =  (a  +  b)(a  +  b);  also,a3  +  3«2/ 
+  3  a  62  _|_  63  _  (a  -f  6)3  r=  (a  +  6)  (a  +  b)  (a  +  b). 

2.  The  difference  between  the  second  powers  of  two  quantitie. 


I 


XVI  PRIME    FACTORS.  69 

may  be  separated  into  two  factors,  one  of  which  is  the  sum  and 
the  other  the  difference  of  the  two  quantities.  For  example, 
a2  _  62  =  (a  +  h)  {a  —  b),  and  x^  _y4  _  (^^^  _|_  ^2)  (x2_y2) 
—  (a;2  -I-  y2)  (a;  _|_  y)  (a;  —  z/).  Also,  25  a^  x^  _  36  6^  c«  = 
(5  a  x2  4-  6  62  c4)  (5  a  x2  —  6  62  c^). 

3.  The  difference  between  similar  powers  of  two  quantities, 
can  be  separated  into  two  factors,  one  of  which  is  the  difference 
of  those  quantities  (Art.  46).  Thus,  a^  —  62^  „3  —  ^3,  a^  —  b'^^ 
qo  —  fj5^  jS^c.  are  all  divisible  by  a  —  b. 

4.  The  difference  between  similar  even  powers  of  two  quanti- 
ties, the  powers  being  above  the  second,  can  always  be  separated 
into  at  least  three  factors,  one  of  which  is  the  sum  and  another 
the  difference  of  the  two  quantities  (Art.  33).     Thus,  x^  —  y'^  = 

5.  The  sum  of  similar  odd  powers  of  two  quantities,  may  be 
separated  into  two  factors,  one  of  which  is  the  sum  of  the  quan- 
tities (Art.  46). 

Thus,  z5  .^  y5  zr  (x  -|-  y)  (x*  —  2;3  y  -|-  x2  y9  —  X  y3  _|_  y4 J 
The  quantity  x^  —  i/^  can  be  separated  into  four  factors.    For, 

a;6_y6_(a;3^^3)(2;3_^3).     But,  2;3-|-y3_(^2_3;^_|_^2)x 

(x  +  y) ;  and,  x^  —  y^  z=  {x^  -{- x  y -^  y^)  {x  —  y).  Hence, 
a;6  _  ^6  —  (2:2  __xy-];-y^){x-{-y)  {x^  J^xy+y'^)  (x  —  y). 

We  have  shown  how  to  find  the  prime  factors  of  simple  quan- 
tities, and,  "in  some  cases,  those  of  polynomials ;  but  any  quan- 
tity which  will  exactly  divide  another,  is  a  factor  of  this  last,  so 
that  some  quantities  may  be  separated  into  factors,  not  prime, 
in  several  different  ways.  For  example,  a^  b  c  -{-  a  b^  c^  =. 
a{abc  +  b^c^)  =  a  6  («  c  +  6  c2)  =  6  (a2  c  +  a  6  c9)  = 
be  'a^-\-abc)=zabc{a-{-bc). 

Art  56.  Sometimes  it  is  desirable  to  find  all  the  divisors  ot 
a  number  or  of  an  algebraic  quantity.  To  explain  the  process 
by  which  this  may  be  effected,  let  us  take  the  quantity  a'^  b^.  If 
we  include  unity  and  the  quantity  itself  among  the  divisors,  it  is 
evident  that  a^  b^  is  divisible  not  only  by  1,  a  and  a^,  but  also  by 
ill  the  combinations  of  these  with  the  different  powers  of  6,  from 


70  DI/ISOKS    OF    QUANTITIES.  XVI 

the  1st  to  the  3d  inclusive ;  that  is,  each  of  the  factors,  1,  a,  aiid 
«2,  is  to  be  taken  once,  h  times,  fe^  times,  and  h^  times.  Hence^ 
if  1  -f- «  +  «^  be  multiplied  by  \ -\- h -\- Ifi -\- h^ ^  which  is  ex- 
pressed thus,  ( 1  -|-  «  +  «^)  ( 1  +  6  +  6^  +  ^^) ,  the  different  terms 
of  the  product  will  constitute  all  the  divisors  of  a^  b^. 

Moreover,  as  no  two  terms  of  this  product  can  be  similar,  the 
number  of  them  will  be  equal  to  the  product  of  the  number  of 
terms  in  one  factor  by  the  number  of  terms  in  thf»  other.  There 
being  3  terms  in  one  factor  and  4  in  the  other,  the  product  will 
contain  3 .  4  or  12  terms.  Performing  the  multiplication,  we 
find  the  following  quantities  for  divisors,  viz  :  1,  c  a^fb,<ab,  a^  6, 
b\ab^,a^b^,b^,ab^,a^b\ 

If  however  any  one  of  the  prime  factors  is  a  pe'ynomial,  all  its 
terms  are  to  be  considered  as  a  single  quantity  m  forming  the 
divisors. 

Art.  57.  We  see,  therefore,  that  the  numbe.  of  divisors  of 
any  quantity  may  be  found,  by  adding  1  to  the  ciponent  of  each 
of  its  'prime  factors,  and  multiplying  these  sums  together  ;  also, 
that  the  divisors  themselves  will  be  the  different  ter  ns  of  the  pro- 
duct, (l  +  a  +  a9  + +  a»)(l  +  b  +  b2+    ....  +  b''')X 

(1  -\-  c -\-  c^ -\- . . . -\-  c"")  Sf'c*  a,  b,  c  ^c.  being  the  prime  fac- 
tors of  the  quantity  in  question,  and  n,  n',  n"  ^c,  t\eir  exponents 

For  example,  let  the  divisors  of  A^w^x^y  or  '^'^m^x^y  be  re 
quired.  The  number  of  divisors  is  equal  to  (2-}'  I)  (3  +  1)  X 
(2  +  1)  (1  +  I)  =  3  .  4  .  3  .  2  =  72  ;  and  the  i^ivisors  them 
Bftlves  will  be  the  terms,  obtained  by  developing  the  product 
(l_^2  +  4)(l  +  m  +  m9  +  m3)  (l  +  x  +  xS)  (l+y). 

Let  the  divisors  o^Za^b-\-Qa^c  be  required.  Now,  3  «2  5  -j, 
6  «2  c  :=  3  a^  (6  -|-  2  c) ;  hence,  the  number  of  divisors  \^  equal 
to  2  .  3 .  2  =  12.  The  divisors  will  be  the  terms  of  /l-^-  3)  X 
(l-|-a  +  a2)(14-6  +  2c);  or  1,3,  a,  3  a,  a^,  3  flS,  (6-|-2c), 
3(6+2c),«(6H-2c),3a(6  +  2c),a2(fe  +  2c),3a2(6  +  2c) 


» It  is  to  be  observed  that  n'  is  read  n  prime,  n"  is  read  n  second,  &c.  The 
accents,  ah  these  marks  are  called,  are  used  merely  to  enable  us  to  represent 
Iffereut  quantities  by  the  same  letter. 


XVII.  GREATEST    COMMON    DIVISOR.  71 

As  an  example  in  numbers,-let  the  divisors  of  160  be  required. 
Separating  it  into  prime  factors,  160  =  2.2.2.2.2.5=: 
2^  .  5  ;  therefore  the  number  of  divisors  is  6  .  2  =  12  ;  and  the 
divisors  are  the  terms  of  (1  +  2  +  4  +  8  +  16  +  32)  (1  -f-  5), 
oi  1,2,  4,  8,  16,  32,  5,  10,  20,  40,  80,  160, 

Let  the  learner  find  the  divisors  of  the  following  quantities. 

1.  a2  62,  5.  150. 

2.  a2  62c3.  6.  780. 

3.  9z2y3.  7.  ^a^b  +  lQab. 

4.  50m2  8.  I0ab  +  25am. 


SECTION  XVII, 


GRXATEST    COMMON   DIVISOR. 


Art.  58.  Any  quantity  which  will  exactly  divide  two  or 
more  quantities,  is  called  a  common  divisor  of  these  quantities ; 
and  the  greatest  that  will  so  divide  them,  is  called  their  greatest 
common  divisor. 

Suppose  A  and  B  two  quantities,  of  which  we  wish  to  find  the 
greatest  common  divisor,  A  being  greater  than  B, 

The  learner  must  remember  that  A  and  B  are  merely  concise 
representations  of  any  two  quantities,  such  as  are  given  in  either 
of  the  examples  succeeding  this  explanation. 

First  divide  A  by  B,  and  if  it  gives  an  exact  quotient,  B  itself 

must  be  the  greatest  common  divisor  ;  for  no  quantity  can  have 

a  divisor  greater  than  itself     But  if  B  will  not  divide  A  exactly, 

suppose  that  we  obtain  a  quotient  Q  and  a  remainder  R.     Then, 

A  =  BQ  +  R. 

For  the  dividend  must  be  equal  to  the  product  of  the  divisor 
pnd  quotient,  plus  the  remainder.     By  transposition, 
Rz=A^BQ, 

Now  any  quantity  which  will  divide  B,  must  divide  Q  times 
B ;  hence,  any  quantity  which  will  divide  A  and  B,  must  exact* 


72  GREATEST    COMMON    DIVISOR.  XVII 

'y  divide  A  —  B  Q,  and  consequently  it  must  divide  R  exactly^ 
since  Rz=zA  —  B  Q.  We  see,  then,  that  B  and  R  will  have 
the  same  common  divisor  as  A  and  B. 

Dividing  now  B  by  R^  if  there  is  a  remainder  R\  it  may  be 
shown  as  before,  that  R  and  R'  will  have  the  samcj  common  di- 
visor as  B  and  22,  that  is,  the  same  as  A  and  B. 

If  we  continue  to  divide  the  preceding  divisor  by  the  remain- 
der, we  shall,  provided  the  given  quantities  have  a  common  di- 
visor, eventually  obtain  a  remainder,  which  will  exactly  di\ide 
the  preceding  divisor,  and  which,  consequently,  must  be  the 
greatest  common  divisor  of  itself  and  that  divisor ;  that  is,  the 
greatest  common  divisor  of  A  and  B. 

Art.  59.    We  have  therefore  the  following 

RULE     FOR     FINDING     THE      GREATEST      COMMON      DIVISOR      OF      TWO 
QUANTITIES. 

Divide  the  greater  hy  the  less,  and  if  there  is  no  remainder,  the 
/«<?s  quantity  will  be  the  divisor  sought;  but  if  there  is  a  remain- 
der, divide  the  first  divisor  by  it,  and  continue  thus  to  make  the 
preceding  divisor  the  dividend,  and  the  remainder  the  divisor, 
until  a  remainder  is  obtained,  which  will  exactly  divide  the  pre- 
ceding divisor ;  this  last  remainder  will  be  the  greatest  common 
divisor  required. 

If  there  are  several  quantities  whose  greatest  common  divisor 
is  required ;  first  find  the  greatest  common  divisor  of  two  of 
them ;  then  find  the  greatest  common  divisor  of  this  divisor  and 
one  of  the  other  quantities,  and  so  on ;  the  last  divisor  thus 
found  will  be  the  greatest  common  divisor  sought. 

The  greatest  common  divisor  of  monomials,  as  well  as  tliat  of 
some  polynomials,  may  frequently  be  found  by  inspection. 

The  application  of  the  preceding  rule  to  polynomials,  some- 
times kads  to  complicated  operations,  unsuited  to  an  elementary 
treutise.     We  shall,  however,  give  a  few  examples. 

It  is  to  be  remarked,  that,  if  one  of  the  quantities  under  con- 
sideration, have  a  factor  not  found  in  the  other,  this  factor  may 
be  left  out  without  affecting  the  common  divisor ;  because  this 
divisor  can  have  no  factors,  not  common  to  both  the  quantities. 


XV 11.  GREATEST    COMMON    DIVISOR.  73 

Moreover,  one  of  the  quantities  may  be  multiplied  by  ai.y 
quantity,  which  is  not  a  factor  of  the  other,  and  which  does  not 
contain  a  factor  of  the  other.  This  is  often  necessary,  in  order 
to  render  the  division  possible.  Thus  a,  which  is  the  greatest 
common  divisor  of  a  m  and  a  d,  is  also  that  of  a  m  c  and  a  d. 

Let  it  be  required  to  find  the  greatest  common  divisor  of  6  «' 

—  6a2y  _|_ 2  03^2 _2y3  and  i2a^—15ai/  +  Sy^. 

We  see  that  2  is  a  factor  of  the  first,  but  not  of  the  second 
and,  that  3  is  a  factor  of  the  second,  but  not  of  the  first.     Leav- 
ing out  these  factors,  we  proceed  to  divide  Sa^  —  Sa^y-f-a^^ 

—  y3  by  4a2  —  Say +y2. 

Operation. 

3  a3  —  ^  a^  y  -\'  a  y^  ^-^  y^.    Multiply  by  4  to  render  the  1st  term 

4  [divisible  by  4  a^. 
12a3-12aay  +  4«y2-4y3|^^'-^^_^y  +  y^^ 

12q3— 15a3y4-3ay2 

Sa^y  -\-  ay^  —  4  y^.    Divide  by  y,  because  it  is  not  a 

[factor  of  the  divisor. 
3  a^  _|_  «  y  —  4  y2^       Multiply  by  4  to  render  division 

4 [possible. 

12a2_[_4a!y_16y2 

12q2  — 15ay  +  3y2 

19  ay  — 19y2.     Divide  by  19 y. 
a  —  y. 

We  now  make  a  —  y  the  new  divisor  and  4a^  —  Say-\-^* 
he  dividend. 


4  a2  —  4  a  y 


—  ay  +  y2 

0. 
Hence  a  —  y  is  the  greatest  common  divisor  sought. 


74  LEAST    COMMON    MULTIPLE  XVID 

Fiud  the  greatest  common  divisors  in  the  fcUowing  examples 

1.  4«2  6c2and  Ua^b^c. 

2.  75a6c2and35a3a;y. 

3.  210  a3,  375  a  X  y  and  45  a  xK 

4.  m^  —  1  and  m^  -\-  nd 

5.  x2__|  andx3  +  l. 

6.  4x4^16x2  and92  +  18. 

7  22:3  — 16x  — 6and3x3_24x  — 9. 

8  6a2+iia2;  +  3x2  and  6«2  +  7aa;__3x2. 

9.  a3_^26_|_3«52_353anda2_«5«6_j_462 


SECTION   XVIIl. 


LEAST    COMMON    MULTIPLE: 


Art.  OO.  When  one  quantity  can  be  exactly  divided  by  an- 
other, the  former  is  called  a  multiple  of  the  Matter.  A  common 
multiple  of  two  or  more  quantities,  is  one  which  is  divisible  by 
them  all ;  and  the  least  common  multiple  is  the  least  quantity 
divisible  by  them  all. 

Suppose  that  it  is  required  to  find  the  least  common  multiple 
of  4  o^  63  and  6  a^  h.  It  is  evident  that  the  quantity  sought  must 
contain  all  the  factors  of  each  of  the  given  quantities.  S« ^para- 
ting  these  into  prime  factors,  ^c^h^  z=.St  .^(v^h^  zn^  cP-  b^,  and 
6  a'*  6  =r  2  .  3  a^  6.  The  different  prime  factors  are  2,  3,  a  and 
6,  and  the  multiple  required  must  contain  as  a  factor  each  of 
these,  as  many  times  as  it  is  found  in  either  of  the  given  quanti- 
ties ;  that  is,  it  must  contain  2  twice,  3  once,  a  four  times,  and 
b  three  times  as  a  factor.  Consequently,  2  .  2  .  3  a^  63  or  12  a^  b^ 
will  be  the  least  common  multiple  required,  for  it  is  manifestly 
ihe  least  quantity  divisible  by  4  a^  b^  and  6  a'*  b. 

Art.  Gl.  Hence  we  deduce  the  following 

RULE     FOR     FINDING    THE     LEAST    COMMON    MULTIPLE    OF   SEVERAIt 
QUANTITIES, 

JF^rst  feparate  the  quantities  into  their  priiu*  factors ;  then 


XIX.  REDUCTION    OF    FRACTIONS.  75 

collect  into  one  product  all  these  different  factors,  each  raised  t6 
the  highest  power  found  in  either  of  the  quantities. 

Required  the  least  common  multiple  in  each  of  the  following 
examples. 

t     a25c,  2a262c,  4a3c.     Ans.  2  .2a^b^c,=zAa^l^c. 

2.     ^m'^y.Qa^m  y^,  12  a^  m^y\ 

3      25a^mx^,  15 a m^ x,  SO a^ m. 

4.  18,  6z2y3^  12a;y^9ax2. 

5.  4a6  +  2ac,  12a6. 

6.  9m2+ 18^32;,  27mx4. 

7.  a^  —  b\a^  +  2ab  +  b^ 

8.  a3__53^a2_62. 

Remark,  It  might  be  proved  that  the  least  common  multiple 
of  two  quantities,  is  equal  to  their  product  divided  by  their  great- 
est common  divisor. 


SECTION    XIX. 

REDUCTION   OF    FRACTIONS    TO   THEIR   LOWEST  TERMS. 

Art.  63.  If  both  numerator  and  denominator  of  a  fraction  be 
aiultiplied  by  the  same  quantity,  the  value  of  the  fraction  will  not 
be  changed;  for,  multiplying  the  numerator  multiplies  the  frac- 
tion, and  multiplying  the  denominator  divides  the  fraction  ;  but 
if  a  quantity  be  multiplied,  ahd  the  product  be  divided  by  the 
multiplier,  the  value  will  remain  the  same  as  at  first. 

Also,  if  the  numerator  and  denominator  of  a  fraction  be  divi- 
ded by  the  same  quantity,  the  value  of  the  fraction  will  not  be 
changed ;  for,  dividing  the  numerator  divides  the  fraction,  and 
dividing  the  denominator  multiplies  the  fraction ;  but  if  a  quan- 
tity be  dividedj  and  the  quotient  be  multiplied  by  the  divisorj  the 
ralue  will  remain  the  same  as  at  first. 

Art.  63.  From  the  principle  last  stated,  we  derive  the  frJ- 
OMing 


76  REDUCTION    OF    FACTIONS.  XIX 


RULiE    FOR    REOrCINO   A    FRACTION    TO    ITS    LOWEST    TERMS. 

Divide  both  numerator  and  denominator  hy  their  greatest  ccm 
mon  divisor. 

Reduce  the  following  fractions  to  their  lowest  terms. 
Zab  b  amx-\-'^5x'^ 

''  WTc    ^"'-  Tc  ^'        b7x^      • 

9x^y^  Zx^y'^+Uxy 

'     l^xy^  21x3  3/3       • 

^      ^la'^bc  ^         125a5  6c2 


155xQm^3/2  2a6  +  13amQ 

7bx^my*'  '  lUa^ 

lUm^xy  Sa^x 


1 728  m3  x^y^  9  a3  ^s  —  6  a^  x 

2^amz  99a^m^ 

"•     ^  ■>. :; t: — •  1  -«. 


231  a^  m2  x  '    33  a3  —  66  a  m^' 

In  the  preceding  examples  the  greatest  common  divisors  of  the 

numerators  and  denominators  are  monomials ;    in  those  which 

follow,  the  greatest  common  divisors  are  polynomials,  but  the 

quantities  may  be  easily  separated  into  factors,  so  as  to  ^exhibit 

the  common  divisors  (Art.  55). 

4a2_462  .  4(«2-_62) 

13.    — — -— -.      This    IS   the   same    as   —-. ~- 

3a  —  36  3  (a  —  6) 

ija  —  b)  (a  +  b)  _  4 (a +  6)  __  ^a  +  Ab 
■      3(a_5)         —         3        —         3 


j^         Sx^y  +  Sxy^  ^^ 


i^  —  x^ 


^x^+Gxy  +  Sy^  a^-^'Zax  +  x^ 

a2_2a6  +  62*  16(23_|_y3)- 

5a3-f.l0a26_|.5q^2  45  a^  63  (a; -f  y ) 

8«34_8a26         •       1^-    25a6(x4  — y4y 

Art.  64.  The  actual  division  of  one  monomial  by  another  is 
impossible,  when  the  coefficient  of  the  former-is  not  divisible  by 
'hat  of  the  latter,  when  the  divisor  contains  any  letter  not  found 


ax  MULTIPLICATION    OF    FRACTIONS    BY    FRACTIONS.  71 

in  the  dividend,  or  when  the  exponent  of  any  letter  in  the  divisoi 
exceeds  the  exponent  of  the  same  letter  in  the  dividend.  A  pol- 
ynomial cannot  be  divided  by  a  monomial,  unless  the  latter  will 
divide  every  term  of  the  former ;  and  the  complete  division  of 
one  entire  polynomial  by  another  is  impossible,  whenever  the 
first  term  of  the  original  or  of  any  partial  dividend,  cannot  be 
divided  by  the  first  term  of  the  divisor,  the  quantities  being 
arranged  according  to  the  powers  of  the  same  letter. 

In  such  cases  the  division  is  expressed  in  the  form  of  a  frac- 
tion, the  divisor  being  placed  under  the  dividend.  The  result 
should  then  be  reduced  to  its  lowest  terms. 

With-  regard  to  polynomials,  however,  we  sometimes  partially 
execute  the  division,  placing  the  last  remainder  over  the  divisor^ 
and  annexing  it  to  the  entire  part  of  the  quotient. 

Express  the  division  in  the  following  examples  and  reduce  the 
results. 

o  ^ 

1.  Divide  Sa^b  hy  lab,     Ans.  — -. 

2.  Divide  13  2;  y  by  26  a  6  c. 

3.  Divide  33  x  y^  by  66  x^  y^. 

4.  Divide  45  a  6  c  by  180  a  6  x\ 

5.  Divide  abc  hy  m-\-n, 

6.  Divide  46  x2  by  4  x^  y  _[.  6  a:  y^. 

7.  Divide  3  z  +  3  y  by  6  (z^  _  y'^y 

8.  Divide  12 1^  +  12  y'^  by  36  x^  +  36  yK 

9.  Divide  13  {x  —  y)  by  39  (xS  —  y^). 


SECTION    XX 


MULTIPLICATION    OF    FRACTIONS    BY    FRACTIONS. 

Art.65.    Whatisfof^?     ^  of  |-=:  g^,  and  §  of  J  =  |> 

ft  /•  I  re 

Wha   is  the  -  part  of  -  1     The  r-  part  of  -  =  7-, ,  and  the 
o  a  b  ^  d        bd 

7* 


78  MULTIPLICATION    OF    FRACTIONS    BY    FRACTIONS.  XX 


a  .  c       ac      .      .     a      c        ac        rii 

T  part  of  -5  =  r— ,;  that  is,  v   •   j  =  i—r       ^n    like    manner 

0  a       oa  o      a       oa 

a      c       m        acm 

b    '  d   '   n        b  d  n 

Hence  we  deduce  the 

RULE    FOR    MULTIPLYING    FRACTIONS    BY    FRACTIONS. 

Multiply  all  the  numerators  together  for  a  new  numerator  ^  ana 
all  the  denominators  together  for  a  new  denominator. 

Remark.  As  the  results  should  be  reduced  to  the  lowest  terms, 
It  is  convenient  to  represent  the  operation  and  omit  the  common 
factors,  previous  to  the  actual  performance  of  the  multiplication 

Thus   -5^      ^+^  -  ^^^("^  +  ^)  -  ±      %\so 

3a        14        86  3.14.8«6  1.2.8  16 


7     •  56»i  *  9a3—  7.5.9a36m""  1.5.  ^a^m"  l^a^m* 
Another  mode  of  proceeding  is  the  following,  viz :  when  frac- 
tions are  to  be  multiplied  together,  if  the  numerator  of  one  and 
the  denominator  of  another  have  common  factors,  omit  those 
factors  before  multiplying. 

36  a2  ft  25  2  y3 

Thus,  if  — —r —  is  to  be  multiplied  by  — ; — ^,  divide  the 
5x2y  *^  16ao3c 

numerator  of  the  first  and  the  denominator  of  the  second  by 
4  a  6  J  also,  the  numerator  of  the  second  and  the  denominator  ot 

the  first  by  5  X  V.     The  fractions  then  become  —  and  ttk~9  ^Le 

^    ,.  ^  .    45ay2 

product  of  which  is  .  -^      . 

4  o^  c  X 

.     ,T  1  .  ,    26c  ^    6a63 

1    Multiply—         ^  ^^"tW 

2.  Multiply  ^  by  -^. 

3.  Multiply  —1^-  by  ^^-p^. 

4.  Multiply —-i—  by—. 


XXI.  ADDiriON    AND    SUBTRACTION    OF   FRACTIONS 


5.   Multipy  -^^  by^-p-^. 

3x2  — 4a;  ^  -         7^ 


6..  Multiply -jj^  by2^3_3^. 

^-  •^"'"P'y  5j=To        ''^  -2r- 

8.  Find  the  product  of  -^ ,  ^^  and  ^^. 

^    TV    1   ,  :.         ^  44-2;     3w2        ^  a4-6 

9,  Find  the  product  of  — ^-^-,  — ^-7  and  -r— J — . 

'^  y3'a_j_5  44-x 

.«    T.    ..   1.  J         /.  39x2  a2__2;2  49«6c 

10.  find  the  product  of  -zr-x-i  -^tk —  and  ; . 

^  7a2*     13x  a-j-x 

11.  Find  the  product  of  ,  — '    ^  „  and  -5 5 

^  9m3  16  a^  a2 — ^9 


12.   Find  the  product  of  li^^+i^),  ^-!z^'  and  -^ 


2m3 


X— y     '      »i2  S{x+y) 


SECTION   XXI, 


ADDITION     AND     SUBTRACTION     OF     FRACTIONS.      COMMON      DENOM- 
INATOR. 

Art.  66.  To  represent  the  addition  and  subtraction  of  frac- 
tions, we  merely  write  them  after  each  other,  with  the  signs  -(- 
and  —  between  them,  being  careful  to  place  these  signs  even 
with  the  line  separating  the  numerator  and  denominator.    Thus, 

tt  C  £ 

-  +  -7  —  -p.     But  if  the  denominators  are  alike,  the  addition 
oaf 

and  subtraction  may  be  performed  upon  the  numerators. 

2  4  2  +  4        6 

Add  together  — ■  and  — -.     Ans.      ,'      =  -— . 
^  11  11  11  11 

4  ^        7        ^        7  —  4        3 
Subtract  ^  from  -.     Ans.  "^  =  jg- 


80  ADDITION    AND    SUBTRACTION    OP    FRACTIONS.  XXI 

All        1     «      ,  ^      *      «4-  6 

Add  together  -  and  -.      Ans.  — - — . 
^  c  c  c 

r^    I  ^    *.  c  .  c  —  6 

Subtract  —  from  — .      Ans.  . 

m  m  m 

Add  together  —  and  -.     Here  tlie   denominators  are  differ* 
0  a 

ent ;  but  if  the  numerator  and  denominator  of  the  first  fraction 

be  multiplied  by  d,  and  the  numerator  and  denominator  of  the 

second  be  multiplied  by  6,  the  denominators  will  be  made  alike, 

without  changing  the  value  of  the  fractions.     The  first  fraction 

becomes  7—,  and  the  second  7-7;  then  adding,  we  have  — ^—-^ 
bd  hd'  ^'  bd 

Ans. 

Add  together  -r ,  -  and  -^.     If  we  multiply  the  numerator  and 

denominator  of  each  fraction  by  the  denominators  of  both  the 

1  1/.       •        ,  adfbcf.bde 

others,  the  fractions  become  7-77.,  ^-f^and  7-7>,    the    sum    of 

o  a  J    oaj         0  a  J 

-.  ,   .    adf-\-bcf-\-bde     .  „.  j    • 

which  IS  — - — '       /..  ' ,  Ans.     H^nce  we  derive  a 

bdf 

RULE    FOR    THE    ADDITION    AND    SUBTRACTIOUT    OF    FRACTIONS. 

Reduce  them  to  a  common  denominator ^  and  then  add  or  sub- 
tract the  numerators. 

Art.  GT.  The  preceding  examples  give  also  the  following  rule 
for  reducing  fractions  to  a  common  denominator. 

Multiply  the  denominators  together  for  a  common  denominator , 
and  multiply  each  numerator  by  all  the  denominators  except  its 
own.  For  this  is  equivalent  to  multiplying  the  numerator  and 
denominator  of  each  fraction  by  the  denominators  of  all  the  oth- 
ers,  which  does  not  alter  the  value  of  the  fractions  (Art.  OS). 

This  rule  for  reducing  fractions  to  a  common  denominator, 
will  uniforndy  give  correct,  but  not  always  the  simplest  results. 

Suppose  it  required  to  reduce  -z — -„  - —  and  - — —   to  a  com< 
'^'^  *  4  m^  6  m  0  m^  x 


XX J.  COMMON    DENOMINATOR.  81 

men  deaoininator.  The  product  of  the  denominators  will,  in 
this  case,  give  a  common  denominator  much  greater  than  is  ne- 
cessary. In  order  to  obtain  the  least,  we  must,  as  in  arithmetic 
find  the  least  common  multiple  of  all  the  denominators  (Art. 
61).  The  least  common  multiple  of  4m3,  6  m  and  3m^x  is 
2^  ^m^x  ;  this  therefore  is  the  least  common  denominator 
sought.  To  produce  this  denominator,  the  first  denominator 
must  be  multiplied  by  3  z,  the  second  by  2  m^  z,  and  the  third 
by  4m;  these,  therefore,  are  the  quantities  by  which  the  nu- 
merators are  respectively  to  be  multiplied.     The  fractions  then 

3ax      'Hbnt^x        .     4cm 
become   —r — -— ,  -— — r—  and 


12wi3x'  I2m^x  12w»3x 

Art.  68.   Hence  we  deduce  a 

RULE  TO  REDUCK  FRACTIONS  TO  THE  LEAST  COMMON  DENOMINATOR. 

JFHnd  the  least  common  multiple  of  all  the  given  denominators, 
and  this  will  be  the  least  common  denominator  sought ;  then  mul- 
tiply  each  numerator  by  the  quantity,  by  which  it  would  be  neceS' 
sary  to  multiply  its  denominator,  in  order  to  produce  the  least 
common  denominator. 

Remark.  The  quantity  by  which  any  numerator  must  be 
multiplied,  may  be  found  by  dividing  the  common  denominator 
by  the  denominator  of  the  given  fraction.  It  is  to  be  observed 
also,  that  fractions  must  be  reduced  to  their  lowest  terms,  before 
we  apply  the  rule  for  reducing  them  to  the  least  common  denom- 
inator. 

1.  AddP-,™  andf. 

q     n  h 

^    ...  3a  2a       .  x 

2.  Add  -— ,  — -  and  -. 

7      5  y 

o    Ajj     4        3x        ,    lly 

3.  Add  -—   — —  and  -— ^. 

7ab   14  a^  21  b^ 

A     *  jj  9x2y    3a6       ^    Ixy^ 
^•^^^■4^'67?""^42P,^ 


i9' 


82 


ADDITION    AND    SUBTRACTICN    OP    FRACTIONS  XX' 

6.  Add  "+4,     "_  and  ^lif-. 

fi    AAA  ^ — 6a-j-*       i« 

^•"^^^   7^'5^'^0T3^^2577»• 
9.  Add—  -^  and-. 

10.  Add  1±^^  and  Izif!. 

1  — x2  l  +  a;2- 

11.  Add  -i—  and  ~. 

l-f-z  I X 

12.  Add  -5^,  and  ^^. 

13.  Add  1^  and        ^"^ 


4  62         4:ab  +  8a 

14.  Subtract  |^  from  — . 

15.  Subtract  ?!=i%romf±^. 

49  28 

16.  Subtract  -^  from  r^^. 

17.  Subtract?^  from  ii+B. 

18.  Subtract  ifi^*  from  ^ 

19.  Subtract  ?|l^^romi±. 

20.. Subtract  =-1^  from  1^. 
7a — 14x  21 

21.  Subtract!^  from  1±^. 

1  +  l2  1  —  x« 


XXII.  DIVISION    BY    FRACTIONS.  BU 


22    Add  — ; —  and  mp. 

In  this  example,  the  integral  quantity  must  be  reducea  to  a 
fraction  having  x-\-y  for  its  denominator. 

23.  Add  a  +  6  and  ^^. 

4 

24.  Add  m2  +  w2  and  ^j^^  +  ^».^, 

'  Im  —  2» 

^K       AAA    1         A    3a2  — 362         - 

^    ^^^  ^  ^^     a2  +  62  ■ 

26.  Subtract      ^^""^     from  3w»« 

27.  Subtract  -^— -  from  5  A3  „3 

x+l 

7  a  6  4-  4  c2 

28.  Subtract  2  from  T       » 

29.  Subtract  a+6  from  ^^  +  ^65^ 

30.  Subtract  »t— 1  from      ^^  . 


SECTION    XXII. 

DIVISIOIV    OF    INTEGRAL    AND     FRACTIONAL     QUANTITIES     BY    FRAC- 
TIONS. 

Art.  69.    How  many  times  is  f  contained  in  8  ?     ^  is  con- 
tained in  8,  40  times,  and  f  is  contained  A^  times. 

How  many  times  is  ^  contained  in  a?   ^  is  contained  in  a,  9 a 

tmes,  and  f  is  contamed  in  a,  -  -  times. 

How  many  times  is  -  contained  in  c  ?    t-   *"  contained  in  c, 
o  0 


64  DIVISION    BY    FRACTIONS.  XXII 


b  c  times,  and  -  is  contained  in  c,  —  times.     This  result  is  th« 
o  a 

same  as  the  product  of  c  by  — 

How  many  times  is  f  contained  in  f  ?     Reduce  the  fractions 

to  a  common  denominator ;  f  =  f  ^  and  f  =::  f  | ;  f  ^  is  contained 

in  f f  as  many  times  as  21  is  contained   in    32,   which  is  ff 

times. 

c  ct 

How  many  times  is  -  contained  in  -  ?     Reduce  the  fractions 
a  0 

,  c        b  c         .   a        ad      b c   . 

to   a   common   denommator ;    —  =  -r~ii  a^"  r  ^^  7— 7  j    7— 7  ^s 

a        bd  b         b  d     bd 

:     , .     ad  .  ,       .  .      ,    . 

contamed  m  — ,   as  many  times  as   be  is   contained   m  /za, 

which  is  ^ — .      This  result   is  the  same  as  the  product  of    - 
be  '^  b 

by-. 

From  the  preceding  questions,  we  deduce  the  following 

RULE    FOR    DIVIDING  AN    INTEGRAL  OR    FRAOTIONAL    QUANTITY    BY 
A    FRAOTJOJf. 

Invert  the  divisor  and  then  proceed  as  in  multiplieation. 

Remark.  When  it  is  required  to  find  what  part  one  quantity 
is  of  another,  make  the  quantity  which  is  called  the  part,  the 
dividend,  and  the  other  the  divisor  :  also,  when  it  is  required  to 
find  the  ratio  of  one  quantity  to  anotner,  make  the  quantity 
which  is  expressed  first,  the  dividend,  and  the  other  the  divisor. 

Perform  the  following  questions,  taking  care  after  invei^ng 
the  divisor  to  omit  common  factors  as  in  Art.  65. 

1.  Divide  m  by  §. 

2.  Divide  a^bc       by  *-y~2' 

3.  Divide  x^  +  y^  hy  j-. 


XXII.  DIVISION    BY    FRACTIONS.  !!i5 

4.  Divide  3(a2  — 62)         by  51?;Zl5) 
5    Divide  —     *  by  4- 


a  X 


6.  Divide  ^  by  -g-r-. 

7.  Divide  — — f  by      .-    _  . 

1  ah  ■'     49  a2 

^    ,-.._,     11x2^0  ^     992;m2n3 

8.  Divide    ^    .  ,/^  by  —. — -rf-, 

9  a^  63  -^       4  a  6* 

r.    T^•  -J    x^-^-^xy  +  if  .      3(x+y) 

10.  Divide  141?  by  ^. 

3  •'       5x 

11.  Divide  — - —  by  — 7—. 

5  4 

.c    Tx.  ..     9x2_3x  ^     z2 

12.  Divide = by  — . 

o  o 

13.  -J is  what  part  of  -77; — r  ? 

4/»  16  wi^ 

..     5x4  «4  35z«5 

14.  r^=— ^      IS  what  part  of  ^.    /  ^? 
17  a  w3  '^  34  a3  w^ 

,^    3(x  +  y).       ^  r9(x2  — y2). 

15.  ^^  "T^^  IS  what  part  of     \^     ^  '  f 

7  x2  *^  14  /» 

,^    „,,     .  \         .      ^76c  +  2162c       1462c2  +  426«, 

16.  What  IS  the  ratio  of ?- to k-^ ? 

4  X  y  9  x"^  y*  »5 

,      „„      .     ,         .      ^155a2  62  65  a  63 

17.  What  IS  the  ratio  of -— r to  ^„    »     ,  _ A 

,0     «ru        .       t-  •         rlO(m+p)  55(m2_«2) 

18.  What  IS  the  ratio  of     ^.^y       to  —^-—-tU  j 

Ibcx"^  56  62  c2  X 

(9.  What  IS  the  ratio  of  —z —  to  r— —  ? 

»i2  —  1  w  -|-  1 

20.  What  IS  the  ratio  of  ^  ^"' +i  ^/ +  ^'^ 

736  c 

12(x3  +  3x2y  +  3xy2  +  y3) 
36562^3 


B6  LITERAL    EQUATIONS.  XX  m 


SECTION    XXIII. 

LITERAL    EQUATIONS. 

Aft.  70.   Let  the  learner  find  the  value  of  z  in  the  following 
equations. 

he  —  z       3x4-4m       ,,,.,. 

a 2</^     ^  I  ^    -     Multiplying  by  a  —  2rf,  we  hav« 

,  ^az-^4tam  —  Qdz — Sdm 

hc  —  z=z ! — . 

o-\-c 

Multiplying  by  6  -|-  c, 

h^c  —  hz-\-hc^ — czz=.Saz-{-4am — 6dz  —  Sdm. 
Transposing  all  the  terms  containing  z  into  the  first  member, 
and  the  others  into  the  second, 

6dz  —  bz  —  cz  —  3  ax  =  4  am  —  b^c  —  bc^  —  Sdm, 

Separating  the  first  member  into  factors,  one  of  which  shall 
be  z, 

(6<Z  — 6  — c  — 3a)a;  =  4am  — 6»c  — 6c2  — 8«fw. 

Here  z  is  taken  6c?  —  b  —  c  —  3 a  times;  that  is,  the  factoi 

6d  —  b  —  c  —  3a  is  the  coefficient  of  z. 

Dividing  by  this  coefficient, 

4  am  —  b^c  —  bc^  —  8  dm 

"^  6rf  — 6  — c  — 3a         '  °^ 

h^c  +  bc^  —  ^am  +  Sdm      -      .    . 

z  = ^— j i— 5 ^-3 I   for  It  is  evident   that 

6  +  c  +  3a  —  Qd 

the  signs  of  both  numerator  and  denominator  may  be  changed 
without  affecting  the  value  of  the  fraction,  since  ^^r-j-  =  +  a, 

and  —  =  -j-  «.     Or  we  might  have  changed  the  signs  of  al 

the  terms  in  the  equation,  previous  to  separating  the  first  mem 
ber  into  factors. 


XXIV                  EQUATIONS    OF   THE    FIRST    DEGREE.  ^   87 

I  —  a        he C% 


2. 


a 


_      a       c  —  X      „ 

3. — -  =  3c  — »i«. 

o  a 


4. 


36  —  4aj cm-^dx 

6  +  c    ""    6  — 2c* 

w^ a;  4-  6  c g-[-4c 

26  —  32  ""       5     • 

6.     ; dc=zox  —  ac. 

0  —  c 

ad^-\-ax^ ac-\-ax 


dx 

d      ' 

ax- 

—  75 

*^  iT 

Ihc  — 

33 

h- 

—  X 

—        2rf 

3a- 

-46 

76- 

-3a 

7- 

-2x 

""     3 

—  X 

SECTION    XXIV. 

■QUATlOUrS    OF   THE   FIRST   DEGREE   WITH   TWO   UirXirOWIT   QUAICTI- 

TIES. 

Art.  71.  The  problems  of  the  first  six  sections  involved  only 
one  unknown  quantity.  When  a  question  involves  several  un- 
known quantities,  there  must  always  be  as  many  conditions 
given,  and,  consequently,  there  must  result  as  many  different 
equations,  as  there  are  unknown  quantities. 

1.  A  man  bought  3  barrels  of  cider  and  2  barrels  of  beer  for 
$14 ;  and,  at  the  same  rate,  5  barrels  of  cider  and  3  barrels  of 
beer  for  $22.     Required  the  price  of  each  per  barrel. 

Let  X  z=z  the  price  of  the  cider  per  barrel, 
and  y  zz:  the  price  of  the  beer  per  barrel. 


P8  EQUATIONS    OF    THE    FIRST    DEGREE  XX1\ 


Tlien,  from  the  conditions  of  the  question, 

(1)  3x  +  2y=14;) 

(2)  5  z  +  3  y  =  22.  )  Multiply  the  1st  by  3  and  the  2d  by  2, 

(3)  92:  +  63^zrr42;) 

(4)  10  z  4-  6  y  =  44.   }  Subtract  the  3d  from  the  4th, 
10z-f~^y  —  9^  —  63^=:  44  —  42;  reduce, 

X  =.  $2.     Substitute  2  for  x  in  the  1st, 
6-|-  2  y  zr:  14 ;  transpose,  reduce  and  divide, 
y  =  $4.     Ans.  Cider  $2  ;  beer  $4  per  barrel. 
We  might  have  multiplied  the  1st  equation  by  5,  the  2d  by  3, 
and  then  subtracted  one  result  from  the  other.    This  would  have 
given  an  equation  without  z,  from  which  we  could  have  found 
the  value  of  y.     Then  if  the  value  of  y  had  been  substituted  in 
one  of  the  preceding  equations,  the  value  of  x  could  have  been 
found. 

2.  A  and  B  together  have  $40,  and  if  3  times  B's  money  be 
subtracted  from  twice  A's,  the  remainder  will  be  $5.  How  much 
money  has  each  ? 

Let  X  z=z  A's  money,  and  y  =  B's.     Then, 

(1)  X   +  y  =  40;) 

(2)  2  z  —  3  y  =    5.   i  Multiply  the  1st  by  3, 
(3J    3  z  +  3  y  =  120  ;  add  the  2d  and  3d, 

5  z  =  125 ;  hence,  x  =.  $25,  A's  money. 
Substituting  25  for  z  in  the  1st, 
25  +  y  =  40  ;  hence,  y  :=  $15,  B's  money. 
We  might  have  multiplied  the  1st  equation  by  2,  and  subtract* 

ed  the  2d  from  the  result,  which  would  have  given  an  equation 

without  z. 

3.  A  market  woman  sold  4  melons  and  6  peaches  for  60  cents, 
and,  at  the  same  rate,  6  melons  and  15  peaches  for  102  cental 
Required  the  price  of  a  melon  and  that  of  a  peach. 

Let  z  r=  the  price  of  a  melon, 

and  y  =  the  price  of  a  peach.     Then, 

(1)  4z+    6y=    f30;> 

(2)  6  z  -h  15  y  =  102.   $ 


X^XIV.  WITH    TWO.  UNKNOWN    QUANTITIES.  89 

The  coefficients  of  y  in  the  two  equations  will  be  alike,  if  the 
1st  be  multiplied  by  5  and  the  2d  by  2;  or  the  coefficients  of  x 
will  be  alike,  if  the  1st  be  multiplied  by  3  and  the  2d  by  2.  But 
the  best  way,  in  this  question,  is  to  divide  the  1st  by  2  and  the 
2d  by  3,  which  gives 

(3)  22;  +  3y  =  30;) 

(4)  2  z  +  5  3^  =  34.   f  Subtract  the  3d  from  the  4th, 
2y  =  4,  and  y  =  2  cents,  price  of  a  peach. 
Substitute  2  for  y  in  the  3d, 

2  X  -j-  6  :=  30,  from  which  xr=  12  cents,  price  of  a  melon. 

4.  Says  A  to  B,  ^  of  my  money  and  $10  is  equal  to  ^  of  yours , 
yes,  says  B,  but  \  of  my  money  and  $10  is  equal  to  |  of  yours. 
H«>w  much  money  has  each  ? 

Let  X  •=.  A's  money,  and  v  ^  B's.     Then, 


Remove  the  denominators, 

(3)  2x4-    60  =  3y;)       _ 

(4)  33^  +  120  =:  8  X.   )  By  transposition, 

(5)  2x  — 3y  =  —    60;  > 

(6)  3  y  —  8  X  =  —  120.   S  Add  the  5th  and  6th, 

—  6  X  =  —  180  ;  change  the  signs, 

6  X  =  180 ;  hence,  x  =  $30,  A's  money. 
Substitute  30  for  x  in  the  3d, 
60  +  60  r=  3y ;  from  which  y  =  $40,  B's  money. 
We  see,  that,  in  the  preceding  problems,  the  conditions,  in 
each  case,  give  rise  to  two  distinct  equations,  which  may  be 
called  the  original  equations ;  the  others  which  follow,  are  de- 
duced from  these,  or  are  mere  modifications  of  them. 

From  the  two   original   equations  containing  two  unknown 

quantities,  we  obtained  one  with  only  one  unknown  quantity. 

This  is  called  eliminating  the  unknown  quantity,  which  this  new 

equation  does  not  contain.     Thus,  in  the  solution  of  the  first 

8* 


00  EQUATIONS    OF    THE    FIRST    DEGREE  XXIV 

question  in  this  article,  we  eliminated  y,  and  thus  obtained  an 
equation  with  x  only  and  known  numbers. 

We  perceive,  moreover,  that  when  the  quantity  to  be  elimina- 
ted, is  in  the  corresponding  members  of  the  two  equations,  that 
is^  o'.*;h^T  in  the  first  or  second  members  of  both,  and  is  found  in 
only  one  term  of  each,  the  following  rule  will,  enable  us  to  effect 
ihe  elimination. 

FIRST    METHOD    OF    ELIMINATION". 

Art.  72.  Multiply  or  divide  the  equations y  if  necessary ,  so  as 
to  make  (he  coefficients  of  the  quantity  to  be  eliminated  the  sanijs 
in  the  two  equations  ;  then  subtract  one  of  the  resulting  equations 
from  the  other ^  if  the  signs  of  the  terms  containing  this  quantity 
are  alike  in  both  equations,  or  add  them  together,  if  the  signs  ai't 
different. 

In  applying  this  rule,  the  equations  should  first  be  freed  from 
fractions,  if  they  .contain  any,  and  it  is  advisable  to  transpose  all 
the  unknown  terms  into  the  first  members ;  moreover,  if  the  un- 
known quantity  to  be  eliminated,  is  found  in  several  terms  in 
one  or  both  of  the  equations,  these  terms,  in  each,  must  be  re- 
duced to  one. 

The  coefficients  of  any  letter  in  the  two  equations  will  be 
made  alike,  if,  after  the  equations  are  prepared  as  prescribed 
above,  each  equation  be  multiplied  by  the  coefficient  of  that  let- 
ter in  the  other  equation ;  or,  if  each  equation  be  multiplied  by 
the  number,  by  which  the  coefficient  of  that  letter  in  this  equa- 
tion must  be  multiplied,  in  order  to  produce  the  least  common 
multiple  of  the  two  coefficients  of  the  letter  to  be  eliminated. 

For  example,  in  the  3d  question,  the  least  common  multiple 
of  4  and  6,  the  coefficients  of  x,  is  12,  which  may  be  produced 
by  multiplying  4  by  3,  or  6  by  2.  If,  therefore,  the  1st  equation 
be  multiplied  by  3  and  the  2d  by  2,  the  coefficients  of  x  will  be 
ftlike. 

1.  A  shoemaker  sold  3  pairs  of  shoes  and  4  pairs  of  boots  for 
$26 ;  and,  at  the  same  rate,  5  pairs  of  shoes  and  3  pairs  of  boots 
♦or  $25.     What  was  the  price  of  the  shoes  and  the  boots  a  pair  \ 


XXIV  WITH    TWO    UNKNOWN    QUANTITIES.  ^  91 

Let  X  =  the  price  ofa  pair  of  shoes, 

and  y  =  the  price  of  a  pair  of  boots.     Then^ 

(1)  32;  +  4y=:26;) 

(2)  5  X  +  3y  =:  25.  )  Transpose  4y  in  1st  and  divide    y  3 

26 4  y 

(3)  z  = — -,     Transpose  3  y  in  2d  and  divide  by  5, 

(4)  x  =  ?^^. 

Now,  since  the  second  members  of  equations  3d  and  4th  aie 
each  equal  to  a;,  they  are  equal  to  each  other.     Hence, 

25  — 3y       26  — 4y      t»,  ,  .  .    ,.    ^      j« 

— -  = 5 — -.     Multiply  by  5  and  3,  or  15, 

o  o 

75  —  9y  =  130  —  20 y;  transpose,  reduce  and  divide, 

y  =  $5,  price  of  a  pair  of  boots. 

Substitute  5  for  y  in  the  3d, 

26—20       ^^       .        ^        ...  1^ 
X  = —  =  f  2,  price  of  a  pair  of  shoes. 

o 

2  What  fraction  is  that  to  the  numerator  of  which  if  4  be 
added,  the  value  of  the  fraction  will  be  ^;  but  if  7  be  added  to 
the  denominator,  the  value  will  be  ^  ? 

Let  X  z=z  the  numerator,  and  y  =  the  denominator. 

X 

The  required  fraction  then  will  be  expressed  by  -     Hence 

X  ft 

(2)  — p--  =  i-    \  Multiply  the  1st  by  y, 

(3)  x-}-4:=.^;  transpose  the  4, 

(4)  a;=|— 4.     Multiply  the  2d  by  y  +  7, 

y-L7 

(5)  X  =  — ^ —      Put  the  two  values  of  x  equal, 
|-4  =  ^±I;  multiply  by  10, 


92  EQUATIONS    OF    THE    FIRST    DEGREE  XXIV 

5y  —  40=z2y-\-i4;  transpose,  reduce  and  divide, 
y  =  18,  the  denominator. 

Substitute  18  for  y  in  the  4th, 
X  =  J/  —  4 1=  5,  the  numerator. 
The  fraction  sought  then  is  -^^. 
In  the  solution  of  the  last  two  questions,  we  found  the  valufl 
of  X  from  each  of  the  original  equations,  as  if  y  were  known ; 
that  is,  we  found  from  each  an  expression  for  z,  consisting  of  y  g 
and  known  numbers ;  then,  by  equalizing  these  two  values  of  x, 
we  obtained  an  equation  without  x.     We  might  have  eliminated 
y  in  a  similar  manner,  and  found  an  equation  without  that.lettei 
Hence,  we  have  a 

SECOND    METHOD    OF    ELIMINATION. 

Arl.  73.  Mnd  the  value  of  one  of  the  unknown  quantities, 
from  each  of  the  equations y  as  if  the  other  unknown  quantity  were 
determined ;  then  form  a  new  equation  hy  putting  these  two  val- 
ues equal  to  each  other. 

Observe,  however,  that  the  unknown  quantity  itself  must  not 
be  contained  in  any  expression  for  its  value. 

1.  Says  A  to  B,  give  me  one  dollar  of  your  money,  and  I  shall 
have  twice  as  much  as  you  will  have  left ;  yes,  says  B,  but  give 
me  one  dollar  of  your  money,  and  I  shall  have  three  times  as 
much  as- you  will  have  left.     How  much  money  has  each? 

Let  X  =.  A's  money,  and  y  =  B's. 

Then  after  B  has  given  A  $1,  A  will  have  a;+  1,  and  B  will 
have  y —  1  dollars.  But  if  A  gives  B  $1,  A  will  have  z  —  1, 
andB,y  +  l. 

Hence,  from  the  conditions  of  the  question, 

(1)  z+l  =  2y-2;) 

(2)  3  z  —  3  =    y-\-l.  (    The  first  equation  gives 

(3)  x  =  2y  — 3. 

Now  this  value  of  z  may  be  put  instead  of  z  in  the  2d  eoua 
ton ;  but  is  z  in  the  2d  is  multiplied  by  3,  we  must  mult  ■   li 


XXIV  WITH    TWO    UNKNOWN    QUANTITIES.  95 

this  value  by  3,  and  the  result,  6y  —  9,  may  be  substituted  'of 
3  z,  in  the  2d,  which  gives 

6y  —  9  —  3  =  y-(~l>  ^''o™  which  we  deduce 

y  =  $2|,  B's  money. 
Substitute  2f  for  y  in  the  3d, 
X  =  5|  —  3  =  $2|,  A*s  money. 
2.  The  mast  of  a  vessel  consists  of  two  parts ;  \  of  the  lower 
part,  added  to  ^  of  the  upper  part,  makes  28  feet ;  and  5  times 
the  lower  part,  diminished  by  6  times  the  upper  part,  is  equal  to 
12  feet.     Required  the  length  of  each  part. 

Let  X  =z  the  lower,  and  y  z=  the  upper  part.     Then, 

(1)      3-  +  f  =  28;> 

2)    Si— 63/=  12.  )     The  1st  gives 
(3)    y  =  168  — 2x. 
Multiply  this  value  of  y  by  6,  and  substitute  the  result  for-  6  y 
m  the  2d ;  but,  as  6y  has  the  sign  — ,  the  value  of  6y  must  be 
subtracted,  that  is,  its  signs  must  be  changed  when  the  substitu- 
tion is  made.     Hence,  we  have 

5x  —  1008  +  12 z  =  12,  which  gives 
a:  =  60  feet,  the  lower  part. 

Substitute  60  for  x  in  the  3d, 
y  =  168  — 120  =  48  feet,  the  upper  part. 
From  the  solution  of  the  foregoing  questions,  we  deduce  a 

THIRD    METHOD    OF    ELIMINATIOIf . 

Art.  74L.  Findy  from  one  of  the  equations^  the  value  of  thf, 
quantity  to  he  eliminated,  as  if  the  other  unknown  quantity  were 
determined,  and  substitute  this  value  in  the  other  equation,  in- 
stead of  the  unknown  quantity  itself 

1.  There  are  two  numbers,  such  that  if  15  times  the  2d  be 
added  to  the  1st,  the  sum  will  be  53;  and  if  3  times  the  1st  be 
added  to  the  2d,  the  sum  will  be  27.  What  are  these  num- 
bers? 


94  EQUATIONS    OP    THE    FIRST    DEGREE  XXIV 

2.  Two  men  talking  of  their  money,  the  first  says  to  the  sec 
ond,  ^  of  mine  and  ^  of  yours  make  $6 ;  but  |  of  mine  and  ^  of 
yours  make  only  $5f .     How  much  money  has  each  ? 

3.  A  farmer  sells  to  one  man  5  cows  and  7  oxen  for  $370 
and  to  another,  at  the  same  rate,  10  cows  and  3  oxen  for  $355 
Required  the  price  of  a  cow  and  that  of  an  ox. 

4.  Says  A  to  B,  give  me  $4  of  your  money,  and  I  shall  have 
as  much  as  you ;  yes,  says  B,  but  give  me  $4  of  your  money, 
and  I  shall  have  three  times  as  much  as  you.  How  much  money 
has  each? 

5.  I  can  buy  in  the  market  3  bushels  of  potatoes  and  4  bush- 
els of  corn  for  $5,  and,  at  the  same  rate,  6  bushels  of  potatoes 
and  7  bushels  of  corn  for  $9.  What  is  the  price  of  a  bushel  of 
each? 

6.  A  man  bought  some  wheat  at  8s.  per  bushel,  and  some  rye 
at  5s.  per  bushel,  to  the  amount  of  $20 ;  he  afterwards  sold,  at 
the  same  rate,  ^  of  his  wheat  and  f  of  his  rye  for  $9^.  How 
many  bushels  of  each  did  he  buy,  and  how  many  of  each  did  he 
sell? 

7.  A  laborer  wrought  8  days,  having  his  son  with  him  b  days, 
and  received  for  both  $10;  he  afterwards  wrought  10  days,  hav- 
ing his  son  with  him  9  days,  and  received  $13.  What  were  the 
daily  wages  of  himself  and  son  ? 

8.  What  fraction  is  that,  whose  numerator  being  dovliled,  and 
the  denominator  increased  by  8,  the  value  becomes  § ;  but  the 
denominator  being  doubled,  and  the  numerator  increv^d  by  2 
the  value  becomes  ^  ? 

9.  What  fraction  is  that,  whose  numerator  being  diminished 
by  3,  the  value  becomes  f ;  but  the  denominator  bei.*g  dimin- 
ished by  3,  the  value  becomes  ^? 

10.  A  man  bought  coffee  at  12  cents  and  tea  at  'to  cents  a 
pound,  and  paid  for  the  whole  $249;  the  next  day  he  disposed  of 
I  of  his  coffee  and  f  of  his  tea  for  $180,  which  was  $10«iO  more 
•ban  it  cost  him.  How  many  pounds  of  each  article  did  he  buy 
and  how  much  of  each  did  he  sell  ? 


XXIV  WITH    TWO    UNKNOWN    QUANTITIES.  95 

1 1.  A  market-woman  bought  eggs,  some  at  two  for  a  cent; 
and  some  at  five  for  three  cents,  and  gave  for  the  whole  f  I  '60 ; 
she  afterwards  sold  them  all  for  $3-10,  and  thereby  gained  ^  a 
cent  on  each  egg.     How  many  of  each  kind  did  she  buy? 

12.  A  grocer  having  two  casks  of  wine,  drew  out  25  ga!?ong 
from  the  smaller  and  30  from  the  larger,  and  found  the  number 
of  gallons  remaining  in  the  former  to  the  number  remaining  in 
the  latter  as  5  to  7;  he  then  put  10  gallons  of  water  into  each 
cask,  and  found  the  number  of  gallons  of  the  mixture  in  the 
smaller  to  the  number  of  gallons  in  the  larger  as  3  to  4.  How 
many  gallons  lid  each  cask  at  first  contain  ? 

13.  A  man  would  sell  4  bushels  of  wheat  and  9  bushels  of 
oats  for  63  shillings ;  or  he  would  exchange  4  bushels  of  wheat 
for  8  bushels  of  oats  and  12  shillings  in  money.  At  what  price 
did  he  estimate  the  wheat  and  oats  per  bushel  ? 

14.  A  grocer  has  two  casks  of  wine,  the  larger  at  12s.  and  the 
smaller  at  10s.  per  gallon,  and  the  whole  is  worth  ,£136 ;  from 
the  larger  he  draws  60  gallons,  and  from  the  smaller  20 ;  he  then 
mixes  the  remainders  together,  adds  40  gallons  of  water  to  the 
mixture,  and  finds  it  worth  9s.  per  gallon.  How  many  gallons 
were  there  at  first  in  each  cask  ? 

15.  A  sportsman  has  a  fishing  rod  consisting-  of  two  parts; 
twice  the  upper  part  exceeds  the  lower  by  5  feet ;  moreover,  3 
times  the  lower  part  added  to  4  times  the  upper  part,  exceeds 
twice  the  whole  length  of  the  rod  by  55  feet.  What  is  the  length 
of  each  part  1 

16.  A  owes  $600,  and  B  $800 ;  but  neither  has  sufficient  to 
pay  his  debts.  Says  A  to  B,  lend  me  ^  of  your  money,  and  I 
shall  be  enabled  to  discharge  my  debts ;  yes,  says  B,  but  lend 
me  i  of  your  money,  and  I  can  discharge  mine.  How  much 
money  has  each  in  possession  ? 


96  'EUUATIONS    OP    THE    FIRST    DEGREE  XXV 


SECTION    XXV. 

CQUATIOirs  I  F  THE  FIRST  DEGREE  WITH  SEVERAL    UM&KOWN   Q.VAN 

TITIES. 

Art.  75.  1.  A  drover  bought  at  one  time  an  ox,  a  cow  and  a 
calf  for  $65 ;  at  another,  and  at  the  same  rate,  two  oxen,  three 
cows  and  a  calf  for  $145;  and,  at  a  third  time,  three  oxen,  two 
cows  and  a  calf  for  $165.  Required  the  price  of  an  ox,  a  cow, 
ind  a  calf. 

Let  X  =.  the  price  of  an  ox, 
y  =  the  price  of  a  cow, 
and  z  =  the  price  of  a  calf.     Then, 

(1)  x+    tj  +  x=   650 

(2)  2x  +  3y  +  z=145;> 

(3)  Sx  +  2y-\-z=l65.  ) 

Here  we  have  three  distinct  equations,  containing  three  dif- 
ferent unknown  quantities;  and  the  first  step  in  the  solution,  is, 
to  deduce  from  them  two  equations  containing  only  two  different 
unknown  quantities.  Let  us  eliminate  z,  that  is,  obtain  two 
equations  without  z.      - 

First  method.  The  coefficients  of  z  being  alike  in  the  three 
equations,  by  subtracting  the  1st  from  the  2d,  also  the  2d  from 
the  3d,  we  have 

(4)  x  +  2y  =  80;) 

(5)  X—    y  =  20.  i 

Since  equations  4th  and  5th  do  not  contain  z,  we  may  obtain 
0  om  them  the  values  of  x  and  y.    Subtract  the  5th  from  the  4th, 
3y  =  60 ;  hence,  y  =  $20,  price  of  a  cow. 
Substitute  20  for  y  in  the  5th, 
X  —  20  =:  20 ;  hence,'  x  :=l  $40,  price  of  an  ox. 
Substitute  the  valu3s  of  a?  and  y  in  the  1st, 
40  -f  20  +  2  =  65 ;  hence,  z  =  $5,  price  of  a  calf. 


XXV  WITH    SEVERAL    UNKNOWN    QUANTITIES  97 


■ o 

=  165.  ) 


I 


Second  method.     Resume  the  original  equations, 

(1)  ^+  y  +  2^  = 

(2)  2x  +  3y  +  z: 

(3)  Sx  +  2y  +  z 
Deduce  the  value  of  z  from  each  equation,  as  if  x  and  y  were 

known, 

(4)  z=z   65  —  X  —  yO 

(5)  2=145— 22  — 3y;> 

(6)  zz=165  — 3x  — 2y.  ) 

Put  equal  to  each  other  the  values  of  z  in  the  4th  and  5th  i 
also,  the  values  of  z  in  the  5th  and  6th, 

(7)  65  —  X  —  yi=145— .2x  — 3y; 

(8)  145— 2x  — 3y  =  165  — 3x  — 2y. 
Transpose  and  reduce  in  the  7th  and  8th, 

(9)     x  +  2y  =  80;) 

(10)  X  —  y  =  20.  / 

Find  the  value  of  x  in  the  9th,  also  in  the  10th, 

(11)  x  =  80-25^;) 

(12)  x  =  20+    7/.  } 

Put  the  values  of  x  in  the  11th  and  12th  equal, 

20  +  3/  =  80  —  2  y  ;  hence,  y  =  $20,  price  of  a  cow. 

Substitute  20  for  t/  in  the  12th, 

X  =  20  +  20  =  $40,  price  of  an  ox. 

Substitute  40  for  x  and  20  for  y  in  the  4th, 
2  =  65  —  40  —  20  =  $5,  price  of  a  calf. 

Third  method.     Take  the  original  equations, 

(1)  ^+    I/  +  ^=    65 

(2)  2x 

(3)  3x 

Deduce  the  value  of  2  from  the  1st,  as  if  x  and  y  were  known 

(4)  2  =  65  —  X  —  y. 
Substitute  this  value  of  z  in  the  2d  and  3d, 

(5)  2x-i-3y-f  65  — X  — y=145; 

(6)  3x  +  2y  +  65  — X  — y=165. 
9 


+  y  +  2=  650 
+  3y  +  2=]45;^ 
+  2y  +  2=165.  ) 


98  EQUATIONS    OF    THE    FIRST    DEGREE  XXV 

Transpose  and  reduce  in  the  5th  and  6th, 

(7)  x  +  2y=:   80;) 

(8)  2x  +  i/=100.  } 
Deduce  the  value  of  x  from  the  7th, 

(9)  x=:80  — 2y. 

Double  this  value  of  z,  and  substitute  the  result  in  the  8th, 

160 — 4y+y=100;  which  gives  y  =  $20,  price  of  a  cow 
Substitute  20  for  y  in  the  9th,  and  we  have 

X  =  $40,  price  of  an  ox. 
Substitute  40  for  x  and  20  for  y  in  the  4th,  and  we  have 

z  =z  $5,  price  of  a  calf. 
2.  There  are  three  men.  A,  B  and  C,  whose  ages  are  such, 
that  iff  of  A's,  ^  of  B's  and  f  of  C's  be  added,  the  sum  will  be 
70  years ;  if  twice  A's  be  added  to  B's,  the  sum  will  be  5  times 
C's ;  and  if  ^  of  A's  be  subtracted  from  ^  of  B's,  the  remainder 
will  be  ^  of  C's.     Required  their  ages. 

Let  ar,  y  and  z  reptesent  the  respective  ages  of  A,  B  and  C. 
Then, 

(1)  ^^  +  i!/  +  i^  =  '70;) 

(2)  2x+     y  =  5z;  > 

(3)  i!/-ix  =  iz.  ) 

Remove  the  denominators  in  the  1st  and  3d,  and  bring  down 
the  2d, 

(4)  16x  +  21y  +  18«=1680 
(2)     ^x  +  y  =  5z; 

(5)  — 2x  +  3yr=3«. 
JFHrst  method.     Transpose  all  the  unknown  quantities  into  the 

first  members,  and  bring  down  the  4th, 

(4)     16x  +  21y  +  18z=1680; 

(6)  2a:  +  y  — 52  =  0; 

(7)  ~-2z  +  3y-^32=:0. 

To  eliminate  x,  add  the  6th  and  7th ;  also,  add  the  4th  to  8 
imies  the  7th, 

(8)  4y  — 8z  =  0;  ) 

(9)  45y  — 6z=1680.  ) 


1 


XXV.  WITH    SEVERAL    UNKNOWN    QUANTITIES.  99 

To  eliminate  z  from  these  two  last  equations,  divide  the  8th 
by  4  and  the  9th  by  3, 

(to)     y-2z  =  0;  ) 

(11)     15y  — 2z  =  560.  i 
Subtract  the  10th  from  the  11th, 

14  y  =  560  ;  hence,  y  =  40  years,  B's  age. 
Substitute  40  for  y  in  the  10th, 

40  —  2  z  z=  0 ;  hence,  z  =  20  years,  C's  age. 
Substitute  40  for  y  and  20  for  z  in  the  2d, 

2  2  -|-  40  =  100 ;  from  which,  x  =  30  years,  A's  age. 
Second  method.     Take  the  original  equations  eleared  of  frac- 
tions, that  is,  the  4th,  2d  and  5th,  inverting  the  order  of  the 
members  in  the  2d  and  5th, 

(4)  16a:  +  21y+l8z=1680;^ 
(2)     52  =  2a:  +  y;  \ 

(5)  32  =  — 22:  +  3y.  ) 
Deduce  the  value  of  z  froin  each  of  these, 

(7)  .  =  '-^; 

(8)  z  =  ^I^^^. 

Put  equal  to  each  other  the  values  of  z  in  the  6ih  and  8th  • 
also,  the  values  of  z  in  the  7th  and  8th, 

—  2a;  +  3y  _  1680— IGz  — 21y 
^^^  3  ■-  18  ' 

(10)     2xj^^zz?i_tl3i. 

Clear  the  9th  and  10th  of  fractions,  transpose,  lednee,  and  find 
»he  value  of  x  from  each, 

(U)     ,^lC80-39y. 

(,2)     x  =  «J! 


I*K)  EQUATIONS    OF    THE    FIRST    DEGREE  XXV 

Put  the  values  of  x  in  the  11th  and  12th  equal, 

3y       1680  — 39  V    , 

=1 ^  J  hence,  y  =  40  years,  B  s  age. 

Substitute  40  for  y  in  the  12th, 

a;  :=  30  years,  A's  age. 

Substitute  30  for  x  and  40  for  y  in  the  7th, 

60  +  40       100       ^^ 
z  = ^ =  — -  =  20  years,  C's  age. 

Third  method.     Resume  the  4th,  2d  and  5th"  equations, 

(4)  16  a;  +  21  y+  18  2  =  1680 ;  ^ 
(2)     5z  =  2x  +  y;  [ 

(5)  32  =  — 22:  +  3y.  ) 
Deduce  the  value  of  z  from  the  5th, 

(6)  .  =  ^^1^.  ■        , 

Substitute  this  value  in  the  4th  and  .2d, 

(7)  16x  +  21y— 12  2+18^  =  1680,0 

(8)  -'«-  +  '^^  =  2.+y.  ■    \ 

Clear  the  8th  of  fractions,  and  reduce  the  7th  and  8th, 

(9)  42  +  39y  =  1680;  ) 
(10)     42  =  3y.  S 

Deduce  the  value  of  x  from  the  10th,  ^ 

(U)    .  =  % 

Substitute  this  value  in  the  9th, 

3  y  +  39  y  =  1680 ;  hence,  y  zr:  40  years,  B*6  a^je. 
Substitute  40  fory  in  the  11th, 

2  =  f  of  40  =  30  years,  A's  age. 
Substitute  30  for  x  and  40  for  y  in  the  6th, 

—  60  +  120       „„      ^^ 

z  = ^ ='A|^  =  20  years,  C  s  age. 


XXV.  WITH    SEVERAL    UNKNOWN    QlTANTITIES.  lOl 

3.  A  boy  bought  of  one  man  3  apples,  2  peaches,  4  pears  and 

2  oranges  for  22  cents ;  of  a  second,  at  the  same  rate,  2  apples, 

3  peaches,  2  pears  and  4  oranges  for  24  cents ;  of  a  third,  5  ap- 
ples, 1  peach,  6  pears  and  10  oranges  for  36  cents;  and  of  a 
fourth,  4  apples,  3  peaches,.  2  pears  and  8  oranges  for  32  cents. 
What  did  he  give  apiece  for  each  1 

Let  X  represent  the  price  of  an  apple,  t/  that  of  a  peachy  z  thit 
of  a  pear,  and  u  that  of  an  orange.     Then, 

(1)  Sx  +  2y  +  Az+   2m  =  22 

(2)  2x  +  Si/  +  2z+   4w  =  24 

(3)  5x+    !/  +  Gz+10u=:S6: 

(4)  42;  +  3y  +  22+   8m  =  32. 
Let  us  eliminate  y,  that  is,  obtain  three  equations  without  y. 

The  three  equations  below  may  be  found  as  follows.  To  obtain 
the  5th,  multiply  the  3d  by  2,  and  subtract  the  1st  from  the  re- 
sult, To  obtain  the  6th,  subtract  the  2d  from  the  4th.  To  ob- 
tain the  7th,  multiply  the  3d  by  3,  and  subtract  the  4th  from  the 
result. 

(5)  7x  +  8z+l8u  =  50: 

(6)  2x  +  4 

(7)  llx+16z  +  : 

Now  let  us  eliminate  z  from  the  last  three  equations.  But 
since  the  6th  does  not  contain  z,  we  may  divide  it  by  2  and  place 
the  result  below.  To  obtain  the  9th,  multiply  the  5th  by  2  and 
subtract  the  7th  from  the  result. 

(8)  x  +  2u  =  ^;       ) 

(9)  Sx  +  14u  =  2i.i 

To  eliminate  x  from  the  last  two  equations,  multiply  the  fiAh 
l>y  3  and  subtract  the  result  from  the  9th. 

.  8  z<  =  12 ;  hence,  u=^l^  cent,  price  of  an  orange. 
Substitute  the  known  value  of  w  in  the  8th,  and  we  have 

z  1=  1  cent,  price  of  an  apple. 
Substi  ,ute  the  values  of  x  and  u  in  the  5th,  and  we  have 
z  zz:  2  cents,  price  of  a  pear. 
9* 


z+18w=:50;     ^ 
u  =  8;  [ 

16z-f-22w  =  76.  ) 


102  EQUATIONS    OF    THE    FIRST    DEGREE  XXV 

Substitute  the  values  of  x,  z  and  m,  in  the  3d,  and  we  have 
y  =  4  cents,  price  of  a  peach. 

Let  the  learner  solve  this  question  by  the  second  and  third 
methods  of  elimination. 

From  the  solution  of  the  foregoing  questions,  it  is  easy  to  see^ 
that  the  three  modes  of  elimination  given  in  the  last  section,  may 
be  extended  to  any  number  of  equations. 

To  apply  the  Jirst  method  for  the  purpose  of  eliminating  a  par- 
ticular quantity  from  several  equations,  it  is  only  necessary  to 
operate  upon  these  equations  taken  two  and  two. 

In  applying  the  second  method  to  several  equations,  we  must 
find,  from  each  equation  that  contains  it,  the  value  of  the  un- 
known quantity  to  be  eliminated,  and  then  put  any  two  of  these 
expressions  for  its  value  equal  to  each  other. 

To  extend  the  third  method,  we  must,  after  having  found  from 
one  of  the  equations  the  value  of  the  unknown  quantity  to  be 
eliminated,  substitute  this  value  in  every  other  equation  that  con- 
tains this  unknown  quantity. 

If  a  question  involves  five  unknown  quantities,  and  gives  rise 
to  five  different  equations,  we  should  first  deduce  from  them  four 
equations  with  only  four  different  unknown  quantities ;  secondly, 
from  these  we  should  deduce  three  equations  with  only  three  un- 
known quantities ;  thirdly,  from  these  three,  two  with  only  two 
unknown  quantities ;  and,  finally,  from  these  two,  one  equation 
with  only  one  unknown  quantity,  from  which  the  value  of  this 
unknown  quantity  might  be  determined. 

Or  if  six  equations  containing  six  unknown  quantities,  were 
given,  we  should  first  obtain  from  them  five  containing  only  five 
unknown  quantities,  and  then  proceed  as  before ;  and  so  on,  if  a 
fitill  greater  number  of  equations  were  given. 

If  either  of  the  equations  does  not  contain  the  unknown  quan- 
tity to  be  eliminated,  that  equation  may  be  put  aside  to  be  placed 
with  the  next  set  of  equations,  viz  :  those  which  contain  one  lesi 
unknowi  quantity. 


XXV.  WITH    SEVERAL    UNKNOWN    QUANTITIES.  103 

Either  of  the  methods  of  elimination  may  be  used,  but  the 
first  will  generally  be  found  most  convenient,  because  it  does  not 
give  rise  to  fractions.  It  will,  however,  be  useful  for  the  learner 
to  perform  every  example  by  each  method,  in  order  to  familiar- 
ize him  with  the  process,  and  enable  him  to  judge  which  will  be 
best  in  any  particular  case.  It  is  not  necessary  that  the  same 
mode  of  elimination  should  be  pursued  throughout  the  solution 
of  a  question,  but  either  of  them  may  be  resorted  to  whenever  it 
shall  seem  the  most  convenient. 

4.  A  merchant  bought  at  one  time  4  barrels  of  flour,  3  barrels 
of  rice,  and  2  boxes  of  sugar  for  $72 ;  at  another,  2  barrels  of 
flour,  5  barrels  of  rice,  and  3  boxes  of  sugar  for  $84 ;  and  at 
a  third  time,  5  barrels  of  flour,  9  barrels  of  rice,  and  8  boxes  of 
Bugar  for  $187.  What  were  the  flour  and  rice  per  barrel,  and 
what  was  the  sugar  per  box  1 

5.  There  are  three  numbers,  such  that  if  3  times  the  2d  be 
subtracted  from  4  times  the  1st,  and  twice  the  3d  be  added  to 
the  remainder,  the  result  will  be  9 ;  if  twice  the  1st  and  5  times 
the  2d  be  added,  and  from  the  sum  3  times  the  3d  be  subtracted, 
the  remainder  will  be  4 ;  and  if  5  times  the  1st  and  6  times  the 
2d  be  added,  and  from  the  sum  twice  the  3d  be  subtracted,  the 
remainder  will  be  18.     What  are  these  numbers? 

6.  Three  boys.  A,  B  and  C,  counting  their  money,  it  was 
found  that  twice  A's  added  to  B's  and  C's,  would  make  $525 ; 
that  if  A's  and  twice  B's  were  added,  and  from  the  sum  C's  were 
subtracted,  the  result  would  be  $3'00 ;  and  the  three  together  had 
$325.     How  much  money  had  each  1 

7.  Three  men  owed  together  a  debt  of  $1000,  but  neither  of 
them  had  sufficient  money  to  pay  the  whole  alone.  The  first 
could  pay  the  whole,  if  the  second  and  third  would  give  him  ^^ 
of  what  they  had ;  the  second  could  pay  it,  if  the  first  and  third 
would  give  him  ^^  of  what  they  had ;  and  the  third  could  pay  it, 
if  the  first  and  second  w^ould  give  him  ^^  of  what  they  had 
Hovv  much  money  had  each  1 


104  EQUATIONS    OF    THE    FIRST    DEGREE.  XXV 

8.  The  ages  of  three  men,  A,  B  and  C,  are  such,  th^t  ^  of 
A's,  I  of  B's  and  ^  of  C's  make  SO  years ;  ^  of  A's,  ^  of  B's  and 

-|  of  G's  make  78  years ;  and  ^  of  A's,  ^  of  B's  and  ^  of  C's 
make  35  years.     Required  the  age  of  each. 

9.  Four  men  could  earn  together  in  one  day  26  shillings.  If 
the  1st  wrought  6  days,  the  2d  8,  the  3d  9,  and  the  4th  12,  they 
would  all  earn  237s. ;  if  the  1st  wrought  2  days,  the  2d  5,  the  3d 
7,  and  the  4th  9,  they  would  earn  161s. ;  and  if  the  1st  wrought 
4  days,  the  2d  3,  the  3d  2,  and  the  4th  1,  they  would  earn  60s 
What  were  the  daily  wages  of  each  ? 

10.  A  merchant  had  four  kinds  of  tea,  marked  A,  B,  C  and 
D.  If  he  mixed  7  pounds  of  A,  10  of  B,  12  of  C  and  18  of  D, 
the  whole  mixture  would  be  worth  $22*90 ;  if  he  mixed  4  pounds 
of  A,  5  of  B,  8  of  C  and  11  of  D,  the  whole  would  be  worth 
^1380;  if  he  mixed  4  pounds  of  A  and  9  of  C,  the  mixture 
would  be  worth  $5*70 ;  and  if  he  mixed  18  pounds  of  A,  12  of 
B  and  36  of  D,  the  mixture  would  be  worth  $31  80.  What  was 
each  kind  worth  a  pound  ? 

11.  I  find  that  I  can  buy  in  the  market  1  bushel  of  wheat,  2 
bushels  of  rye,  3  of  barley,  4  of  oats  and  6  of  potatoes  for  $12 ; 
3  bushels  of  wheat,  4  of  rye,  8  of  barley,  3  of  oats,  and  4  of  po- 
tatoes for  $24^  ;  5  bushels  of  wheat,  2  of  rye,  10  of  barley,  6  of 
oats  and  8  of  potatoes  for  $32 ;  and  8  bushels  of  wheat,  7  of  rye, 
6  of  barley,  5  of  oats  and  4  of  potatoes  for  $35^ ;  moreover,  that 
a  bushel  of  wheat  and  a  bushel  of  oats  cost  as  much  as  a  bushel 
of  rye  and  a  bushel  of  barley.  Required  the  price  of  a  bushel 
of  each. 


KXVI.  SUBSTITUTION    OF    ALGEBRAIC    QUANTITISS.  105 


SECTION    XXVI. 

NUMERICAL    SUBSTITUTION    OF    ALGEBRAIC    QUANTITIES 

Art.  76.    Find  the  numerical  value  of  the  following  quanti- 
163,  when  a=z5,  6  =  6,  c  =  7,  and  d=.  10. 

18.  a  +  bcd. 

19.  2ab  +  Sd. 

20.  a^-^-b  +  c  +  d. 

21.  ab^+cd\ 

22.  a  +  b  +  d^. 

23.  a-^b  —  c. 

24.  a  +  b  —  c  —  d. 

25.  a^—b  —  c  +  (^. 

26.  (a-^b  +  c)d. 

27.  lb  +  c  +  d)a. 

28.  ab(c-{-d). 

29.  (a  +  *)(c  +  ^)- 

30.  {a  +  b)(d-'c), 

31.  (a  +  6)(c  — £/). 

32.  (a  — fe)(c  — /). 

33.  (62_a2)(c  +  j). 

34.  (a  +  6)(c2  +  rf2). 

35.  (62_a2)(rf2__c2) 

36.  («2_52)(c2__^^. 

37.  (a+bf. 

38.  (a-|-6)2c(/. 
17.   abc  +  d. 

Find  the  value  of  the  following   expressions,  when  a  =  ^ 
t  :=  2,  w  =  4,  and  w  =  6 


1. 

ab,     Ans.  6. 

5: 

=  30. 

2 

a2  6c.  Ans.52 

.6 

.7  =  1050. 

3. 

abed. 

4. 

a^bcd. 

5. 

ab^cd. 

6. 

a^b\ 

7. 

acd^. 

8. 

ab 

9. 

abc 
d  • 

10. 

e^ 
a 

U. 

a^b^ 
c^d' 

12. 

a 
bd' 

13. 

a  +  b-\-c. 

14. 

«6  +  c. 

15. 

a  +  bc. 

_ 

16. 

ab-\-cd. 

39. 

a-\-b — n  +  wi 

42.  6a- 

-5abn. 

40. 

3a  +  26  +  n  — »L 

43.  3m- 

-1  abn. 

41. 

ab-\-mn. 

106  SUBSTITUTION    OF    ALGEBRAIC    QUANTITIES.  XXVI 


44. 

a  +  b 

m  —  n 

45. 

-6  +  a 

n  —  m 

AR 

bm 

47. 

48. 


a  —  4« 
m^  —  4  «  6  » 


a_27  63 
49.   {a  +  b){m  —  n)^ 
~^'  G-jr-ma  '         (fii  —  a)2 

Substitute  numbers  in  the  following  equivalent  expressions, 
showing  their  identity  whatever  numbers  are  put  instead  of  the 
letters,  observing  however  to  give  the  same  value  to  the  same 
letter  in  the  two  members  of  an  equation. 

50.  {a-\-b)  c  =  ac-^b  c. 
Suppose  a=z2,  6  =  3,  and  c  =  6. 

Then,  (a  +  6)  c  =  (2  +  3)  6  =  5  .  6  =  30. 
Also,  ac  + 6 c  =  2.  6  +  3.  6  =  12+18  =  30. 

51.  (a+6)(a  — 6)  =  a2_62. 

52.  («  +  *)^  =  «^  ^2  aft +  62. 

53.  (a  —  m)2  =  aa_2am  +  m2. 

54.  a  +  -  = ! — . 

c  c 


55. 


m —  1  '         • 


56.  r— -  =  m3  —  m^-\-m  —  1. 

m  + 1  ' 

57.  J:   =a^^ab  +  b^ 

58.  ^^f"f  =a2  — 2«  +  4. 
a  +  2  ' 

59.  («  +  6)(a  +  c)  =  a2  +  a(6  +  c)+6c. 

60.  (a  +  6)  (a  +  c)  (a  +  rf)  =  a3  _(.  ^2  (J  ^  c  +  (^  4 
9(bc-\-bd-]-cd)-{-bcd. 

61         6       ,    «  — 6_        «2 
"^-  a  +  6"»~      6      —  a6  +  62-   • 
fl  +  6       fl-6_2(«2  +  62) 
'^'  a  — 6'T"a  +  6-     a2__52   • 


XXVII  GENERALIZATION.  107 


^--2a+l  __  a-~_I 


64    «  +  6       «  —  &         4fl6    _ 
a_6       a  +  6       a2  — 62~"    • 


SECTION    XXVII. 

^kAxralizatiox. 

Art.  77.  In  the  problems  of  the  first  six,  as  also  in  those  of 
the  24th  and  25th  sections,  letters  have  been  used  to  represent 
unknown  quantities  only,  and  the  results,  expressed  in  definite 
numbers,  correspond  to  the  particular  questions  only,  from  which 
they  were  derived. 

But  in  pure  algebra,  letters  may  also  represent  known  quan 
tities,  or  they  may  be  used  indefinitely,  and  afterwards  any  num- 
bers may  be  substituted  in  their  place.  Also  the  results  of  pure 
algebra,  which  are  called  ybrmw/te,  show  by  what  operations  they 
were  obtained,  and  furnish  rules  for  the  solution  of  all  questions 
of  the  same  kind. 

1.  Two  men,  A  and  B,  are  to  share  $420,  of  which  B  is  to 
have  3  times  as  much  as  A.     Required  the  share  of  each. 

Here  the  object  is  to  separate  $420  into  two  parts,  such  that 
one  shall  be  3  times  as  great  as  the  other. 

Let  X  =z  A's  share ; 
then,  3  X  =z  B's  share. 
Hence,  a;  -|-  3  x  =  420 ; 

x=  $105,  A's  share; 
3  X  =  $315,  B's  share. 
Instead  of  420,  in  this  question,  put  the  letter  a ;  then  the 
problem  will  be,  to  separate  the  number  a  into  two  parts,  one  of 
which  shal  be  3  times  the  other. 


108  GENERALIZATION.  XX  V^IJ 

Representing  the  shares  as  before,  we  have 
x-\-Sxzz^a.     Hence, 

z  =  2,  A's  share,       J 

o  /  General  ^ormulae. 

Sx=z—,  B's  share.  J 

If  we  now  put  $420  instead  of  a  in  these  formulae,  we  have 

420 
x  =  —  =  $105,  A's  share,         ^ 

„     .jj„  •     >  Particular  answers. 

3x  =  -^- —  =  $315,  B's  share.  ) 

We  perceive,  from  the  general  formulae,  that  one  part  is  a 
fourth,  and  the  other  three-fourths,  of  the  number  to  be  divided, 
without  regard  to  the  particular  value  of  that  number. 

Let  the  learner  put  other  numbers  instead  of  a  in  the  formulae, 
and  find  the  two  parts.  Any  number  divisible  by  4  will  give 
whole  numbers  for  these  parts. 

2.  A  father  left  by  his  will  $4500  to  be  divided  between  his 
son  and  daughter,  with  the  condition  that  the  son  w«s  to  receive 
$500  more  than  the  daughter.     What  was  the  share  of  each  ? 

In  this  problem  it  is  required  to  separate  $4500  into  two  parts, 
such  that  one  shall  exceed  the  other  by  $500.  Instead  of 
$4500,  let  us  suppose  that  the  number  to  be  separated  into  parts 
is  indefinite,  and  that  it  is  represented  by  a ;  also,  that  b  repre- 
sents the  excess  of  the  greater  part  above  the  less.  Then  the 
problem  is,  to  separate  the  number  a  into  two  parts,  such  that 
the  greater  shall  exceed  the  less  by  6. 

Let  X  =  the  less  part ;  then, 

x-\-bz=.  the  greater  part.     Hence, 

x-^x-\-b=:a.     Reduce  and  transpose  6, 

2x=za  —  b;  divide  by  2, 

a       b       a — b     ,    . 
xz=^  —  -=  -^,  the  less  part 


XXVII.  GENERALIZATION.  1(W 

Tc  obtain  the  greater  we  add  h  to  the  less,  and  we  have 

CL        h 
z-\'h=i-  —  -  4-  6.     Change  h  to  halves, 

x  +  h=z-  — --{-  —  ',  reduce, 

•   t       a   ,   h       a-\-h     _ 
1  +  6=-  +  -==  — g— ,  the  greater  part. 

If  we  examine  the  formulae  for  the  two  parts,  and  recollect 
that  a  and  b  may  stand  for  any  numbers,  provided  that  b  is  less 
than  fl,  we  see  that  they  furnish  the  following  rule  for  separating 
a  quantity  into  two  parts. 

The  less  part  is  found  by  subtracting  half  the  excess  of  the 
greater  above  the  less  from  half  the  number  to  be  separated ;  or, 
by  subtracting  the  excess  of  the  greater  above  the  less  from  the 
number  to  be  separated^  and  dividing  the  remainder  by  2. 

The  greater  part  is  found  by  adding  half  the  excess  of  the 
greater  above  the  less  to  half  the  number  to  be  separated;  or,  by 
adding  the  excess  of  the  greater  above  the  less  to  the  number  to  be 
separated,  and  dividing  the  sum  by  2. 

Let  the  learner  separate  each  of  the  following  numbers  into 
I  wo  parts  by  means  of  the  formulae,  or  by  following  the  rule. 

Numbers  to  be  separated.       Excess  of  one  part  over  the  other 

3.  150 30. 

4.  230    . 60. 

5.  1200 120. 

6.  27              ...'.....  5. 

7.  35      ........     .  3. 

8.  70 3. 

9  47^ 13. 

10.  99 33^. 

11.  Separate  a  number  a  into  three  parts,  such  that  the  mean 
shall  exceed  the  least  by  b,  and  the  greatest  shall  exceed  the 
mean  by  c. 

10 


110  GENERALIZATION.  XXVIL 

Let  X  =  the  least  part ; 

then  x-{-b=.  the  mean  part ; 

and  x-\-b  -\-c=:  the  greatest  part. 

Hence,  x-\-x-\-b-\-x-\-b-\-c  =  a.     Reducing, 

^x-\-2b-\-c=^a;  transposing, 

Sx==.a — 2  6  —  c;  dividing, 

a      2b        c       a  —  26  —  c    ,     , 
ajr=- ^— _  = ,  the  least  part. 

Adding  b  to  the  least,  we  have 

,,        a       2b       c     ,   ^       a       26        c,3i 

a    .    h         c        a-\-b  —  c    . 
=  ^  +  ^  —  -^  =  — '— ,  the  mean  part. 

Aiding  c  to  the  mean,  we  have 

,,,  a   .    b        c,  a,   6        c,3c 

x  +  6  +  c=-  +  ---+.=  -+-_-+^ 

a   ,    6    ,   2c       a  +  6  +  2c    ^ 
=  3  +  3-+  3-  =  — '-3-' ,  the  greatest  part. 

Let  the  learner  translate  these  formulae  into  rules,  recollecting 
that  a  represents  the  number  to  be  separated,  6  the  excess  of  the 
mean  above  the  least,  and  c  the  excess  of  the  greatest  above  the 
mean. 

12.  A  man  bought  sugar  at  a  cents,  flour  at  6  cents,  and  co^ 
fee  at  c  cents  per  pound,  and  the  whole  amounted  to  d  cents. 
How  many  pounds  of  each  did  he  buy,  if  he  bought  the  sam« 
quantity  of  each  1 

Let  X  =z  the  number  of  pounds  each. 
Then,  ax-\-bx-\-cx  =  d. 
Separating  the  1st  member  into  factors,  one  of  which  is  x, 
(a -{- b  -\-  c)  X  =z  d.     Dividing  by  the  coefficient  of  x, 

x  =  — ,   -    . — ,  the  number  of  pounds  of  each. 
a-\-b-\-c 

This  formula  may  be  translated  into  the  following  rule,  viz : 
Divide  the  price  of  the  whole  by  the  sum  of  the  prices  of  a  pouna 
gf  each  sort ;  the  quotient  will  be  the  number  of  pounds  of  each 


XXVII.  GENERALIZATION.  11 

If  in  the  formula  we  substitute  the  numbers  7,  6,  10  and  92 
for  a,  6,  c,  and  d  respectively,  we  have 

92 

X  z=z  ^  ,   ^.  ,    ,^  rr:  f 4  =  4  Ibs.,  particular  answer. 
7  +  G+lO       ^^  *^ 

13.  A  farmer  found  he  had  a  times  as  many  cows  as  oxen, 
and  b  times  as  many  sheep  as  cows,  and  that  h.s  whole  stock 
amounted  to  the  number  c.     Required  the  number  of  each. 

Let  X  ==.  the  number  of  oxen  ; 
then  ax=i  the  number  of  cows ; 
and  abx=z  the  number  of  sheep. 
Hence,  x-\-ax-\-abx=ic. 
Here  x  is  taken  1  -f-  a  -f"  ^  ^  times  ;  therefore, 

{l-{'a-\-ab)xz=c.     Dividing  by  the  coefficient  of  x, 

X  =  -—. • r,  number  of  oxen. 

1  ^a-f-ao 

/• 

ax=  —-. , r  X  a,  number  of  cows. 

1  +  a  +  ao 

abx=z  -— ; X  ciby  number  of  sheep 

l  +  «  +  «o 

If  3  be  put  for  a,  4  for  6,  and  128  for  c,  in  these  formulae,  we 

have 

128 
the  number  of  oxen  =  =  ^^  =.  8 ; 

1  — j—  o~[~  1«* 

the  number  of  cows  =  3  . 8  =  24 ; 
the  number  of  sheep  =  4  .  3  .  8  =  96. 

14.  What  will  be  the  particular  answers  in  the  preceding  ques 
lion,  ifa=:5,  6  =  7,  andc  =  369? 

15.  Two  men  had  engaged  to  perform  a  certain  piece  of 
work  ;  the  first  could  do  it  alone  in  a  days,  and  the  second  in  b 
days.     How  long  would  it  take  both  working  together  to  do  it  ? 

Let  X  =z  the  number  of  days  in  which  both  would  do  it. 
Then,  as  the  first  could  do  the  whole  in  a  days,  in  1  day  he 

H'ould   lo  -  of  it ;  and,  as  the  second  could  do  the  whole  in  i 
a 


112  GENERALIZATION  XXVD 


days,  in  1  day  he  would  do  -  of  it;  so  that  both  would  per 

11  X        z 

form  — I-  -  of  it  in  1  day,  and  in  x  days,  -  4-  t-     But  in  i 
a    ^   b  -"  ^  '  a    *   b 

days,  we  have  supposed  that   they  would   perform   the  vvhol« 

Hence, 


7               7 

— |-  T  =  1>  piece  of  work. 

Multiplying  by  a 

;  and  6, 

bx-{-ax=zab; 

or. 

ab 

T  rrr .       A  n 

dividing  by 

b  +  a, 

.     b  +  a      -—' 

From  this  formula  we  derive  the  following  rule  for  any  similar 
case,  in  which  two  workmen  are  employed. 

Divide  the  product  of  the  numbers  expressing  the  times  in 
which  each  would  perform  it^  by  the  sum  of  those  numbers. 

Let  the  learner  find  the  answers  to  the  following  questions,  by 
means  of  the  preceding  formula. 

16.  If  A  could  perform  a  piece  of  work  in  6  days,  and  B  could 
perform  the  same  in  5  days,  how  long  would  it  take  both  together 
to  perform  it  ? 

17.  By  a  pipe  A,  a  certain  cistern  will  be  filled  with  water 
in  5^  hours,  and  by  another  pipe  B,  it  will  be  filled  in  8^  hours; 
in  what  time  will  it  be  filled,  if  water  flow  through  both  pipes  at 
the  same  time  1 

18.  Let  it  be  proposed  to  find  a  rule  for  dividing  the  gam  or 
loss  in  partnership,  or,  as  it  is  commonly  called,  the  rule  of  feU 
Icwship.     First,  take  a  particular  case. 

Three  men  tr.ided  in  company  and  put  in  stock  in  the  follow- 
ing proportions,  viz ;  A  put  in  $5  as  often  as  B  put  in  $3  and 
as  ofteL  as  C  put  in  $2.  The  company  gained  $650.  Required 
(►ach  man's  share  of  the  gain. 


XXVII.  GENERALIZATION.  113 

Let  X  =z  A's  share.     Then,  since  B  furnished  f  as  much  stock 
as  A,  he  must  have  f  as  much  gain ;  therefore, 

— -  =  B's  share.     In  J  ike  manner, 
o 

2  X 

-—  =  C's  share.     Hence, 
o 

x-\----\-~=z 650.     Multiplying  by 5, 

DO 

52_[-3a;-|-2x  =  3250,  or 
10x  =  3250; 

X  =  $325,  A's  share. 

-?  =  $195,  B's  share. 
o 

^  =  $130,  C's  share. 
5 

'i*o  generalize  this  question,  suppose  that  A  put  in  m  as  often 
as  B  put  in  n  and  as  often  as  C  put  in  p  dollars,  and  that  the? 

pained  a  dollars.     Then  B  puts  in  — ,  and  C   —  as  much  as  A. 

Let  X  =r  A's  gain ;  then, 

—  =  B's  gain,  and 

^—  =  C's  gain.     Hence, 

X-] f-  —  =  a.     Multiplying  by  m, 

mx-\-nx-^px=:7na;  or, 

{m-\-n -\-p) x=ima;  dividing  by  the  coefficient  of  2, 

ma  «  ..     , 

-,  or  ni  X  — i J — ,  A's  share. 


m-\-n  -\-p  m-\-n-\-  p 

B's  share  is  -  of  A's ;  —  of  wi  X  — 7 ; —  is  — : ; — 

m  m  m-\-n-\-p      m-f-n  -}-  p 

n 
ind  —  of  it  is  n  times  as  much ;  therefore, 


m 


nx  a  T»,     1 

—  =  71  X  — f i — ,  B  s  share. 

m  m-\-n-\-p 

,    10* 


1 14  GENERALIZATION.  XXVIl 

In  like  manner,  C's  share  is  —  of  A's :  or 

m 

- —  z=zp  X  -^—, ; — ,  C  s  share. 

m  m-f-n-\-p 

By  examining  these  formulae,  we  perceive  that  the  whole  gair 
a  is  divided  by  m-\-n  -{-p,  the  sum  of  the  proportions  of  th€ 
stock  furnished  by  all  the  partners,  and"  that  this  quotient  is 
multiplied  by  m,  A's  proportion,  by  n,  B's  proportion,  and  by  ^, 
C's  proportion  of  the  stock,  to  obtain  their  respective  shares  of 
the  gain.  Hence,  observing  that  a  may  represent  the  loss  as  well 
as  the  gain,  to  find  each  partner's  share  of  gain  or  loss,  we  have 
the  following  rule. 

Divide  the  whole  gain  or  loss  by  the  sum  of  the  proportions  of 
the  stocky  and  multiply  the  quotient  by  each  partner's  proportion. 

This  rule  is  applicable,  whatever  be  the  number  of  partners. 

19.  Suppose  A  put  in  $400,  B  $300,  and  C  $200,  and  that 
they  gained  $450.  By  the  preceding  formulae,  or  by  the  rule, 
what  was  each  partner's  share  of  the  gain  ? 

Remark.  When  the  sums  actually  put  in  are  given,  the  sim- 
plest proportions  of  the  stocks  will  be  found  by  dividing  these 
Bums  by  the  greatest  number  that  will  divide  them  all  without 
any  remainder.  Thus,  400,  300,  and  200  are  all  divisible  by 
100 ;  and  the  quotients,  4,  3,  and  2,  express  the  proportions  of 
the  stock. 

20.  What  would  be  each  man's  loss,  if  A  furnished  $300,  B 
$150,  and  C  $100,  the  entire  loss  being  $99? 

21.  What  would  be  each  man's  share  of  $500  gained,  if  foui 
partners  furnished  respectively  $800,  $600,  $400,  and  $200? 

22.  A  put  in  $200  for  6  months,  B  $150  for  5  months,  and 
C  $300  for  2  months.  They  gained  $27^ ;  what  was  each  man's 
share  of  this  gain  ? 

Remark.  It  is  evident,  that,  if  the  stocks  are  employed  une- 
qual times,  each  partner's  stock,  or  his  proportion  of  the  stock, 
must  be  multiplied  by  the  number  expressing  the  time  during 
whish  it  is  in  trade,  and  that  then  the  proportions  of  these  pro 
Impels  tnusit  be  used. 


XXVII.  GENERALIZATION.  115 

In  general,  known  quantities  are  represented  by  the  first,  and 
unknown  quantities  by  the  last  letters  of  the  alphabet.  But,  in 
some  cases,  it  is  more  convenient  to  use  the  initial  letters  of  the 
names  of  quantities,  whether  known  or  unknown. 

In  the  following  questions  relating  to  simple  interest,  let  p  rep- 
resent the  principal,  r  the  rate,  t  the  time,  i  the  interest,  a  the 
amount,  and  d  the  discount.  In  these  questions,  r  is  supposed 
to  be  a  fraction,  as  '06,  05,  &c.,  according  as  the  rate  is  6  per 
cent.,  5  per  cent.,  &/C.,  and  the  time  is  supposed  to  be  expressed 
in  years  and  fractions  of  a  year. 

23.  What  is  the  simple  interest  of  p  dollars,  for  t  years,  at  r 
per  cent.  ? 

The  principal  multiplied  by  the  rate  gives  the  interest  for  one 
vear;  hence, 

rp^=.  the  interest  for  1  year ;  and 
trpz=.  the  interest  for  t  years.     Therefore, 
i=ztrp. 
This  formula  gives  the  following  rule. 

To  Jind  the  interest  when  the  principal^  ratCy  and  time  art 
known,  multiply  together  the  principal,  time,  and  rate. 

24.  The  principal  being  $256*25,  the  time  4^  years,  and  the 
rate  6  per  cent.,  what  will  be  the  interest  1 

In  the  equation  trp=zi,  provided  any  three  of  the  quanti- 
ties are  given,  the  other  may  be  found.  Let  the  learner  find  the 
formulae  and  make  rules  for  the  following  general  questions,  and 
solve  by  the  rules  the  particular  examples  subjoined. 

25  The  interest,  time,  and  rate  being  given,  to  find  the  prin- 
cipal. 

26.  If  the  interest,  for  7  years  at  5  per  cent.,  is  $26*25,  what 
is  the  principal  ? 

27.  The  interest,  time,  and  principal  given,  to  find  the  rate. 

28.  The  interest  being  $74*4711,  time  6  years,  and  the  prin- 
cipal $225-67,  what  is  the  rate  1 

29.  The  interest,  rate,  and  principal  given,  to  find  the  time. 

30.  If  the  interest  is  $102,  rate  4^  per  cent.,  and  the  principal 
$320,  what  is  the  time  ? 


116  GENERALIZATION  XXVII 

31.  What  is  the  amount  of  p  dollars,  for  t  years,  at  the  rate  r, 
simple  interest  \ 

The  amount  being  the  sum  of  the  principal  and  interest,  we 
have 

a z=p -\-trp;  or,  a  =p  (1  +  < r). 

This  formula  gives  the  following  rule. 

To  Jind  the  ainount^  toJien  the  principal^  timcy  and  rate  art 
J'tiowTiy  multiply  the  time  and  rate  together ^  add  1  to  the  product, 
and  multiply  this  sum  by  the  principal. 

32.  The  principal  being  $650,  rate  4^  per  cent.,  and  the  time 
7  years  and  3  months,  what  is  the  amount  by  the  preceding 
rule? 

The  equation,  p-^-f  rp=::a,  contains  four  different  quanti- 
ties, any  three  of  which  being  known,  the  other  may  be  deter- 
mined. 

Let  the  learner  find  formulae  and  rules  for  the  following  gen- 
eral questions,  and  solve  the  particular  examples  by  the  rules. 

33.  The  amount,  time,  and  rate  being  given,  to  find  the  prin- 
cipal ;  that  is,  to  find  what  sum  of  money  put  at  interest,  at  a 
given  rate,  and  in  a  specified  time,  will  amount  to  a  given  sum. 

N.  B.  The  principal,  in  this  case,  is  sometimes  called  the 
present  worth  of  the  amount. 

34.  What  is  the  present  worth  of  $300,  due  in  3  years  and  4 
months,  the  rate  being  6  per  cent.  ? 

35.  The  amount,  principal,  and  time  given,  to  find  the  rate. 

36.  The  amount  being  $405  09,  principal  $321  50,  the  time  4 
years,  what  is  the  rate? 

37.  The  amount,  principal,  and  rate  given,  to  find  the  time. 

38.  Amount  $352,  principal  $275,  and  rate  8  per  cent.,  re- 
quired the  time. 

39.  The  amount,  time,  and  rate  given,  to  find  the  discount. 
Remark.     The  formula  for  the  discount  may  be  found  by  sub* 

trading  the  formula  for  the  present  worth  from  the  amcmt  a^ 
and  simplifying  the  result 


XXVII  GENERALIZATION.  117 

40  Required  the  discount  on  <£100,  due  in  3  months,  tl  e  rata 
being  5  per  cent. 

41.  At  a  given  rate,  in  what  time  will  a  sum  be  doubled  1  In 
what  time  will  it  we  tripled  ? 

Remark,  Take  the  formula  for  the  amount,  put  2p  and  3^ 
successively  instead  of  a,  and  then  find  the  value  of  t, 

42.  In  what  time  will  a  sum  be  doubled  at  6  per  cent.  ?  In 
what  time  will  it  be  tripled  ? 

43.  In  what  time  will  a  sum  be  doubled  at  5  per  cent.  ?  In 
what  time  will  it  be  tripled? 

44.  Separate  the  number  a  into  two  parts,  one  of  which  shall 
be  n  times  the  other. 

45.  Separate  a  into  two  parts,  so  that  the  second  may  be  the 

-  part  of  the  first. 

n   '^ 

46.  Separate  a  into  three  parts,  such  that  the  second  shall  be 
m  times,  and  the  third  n  times  the  first. 

47.  Separate  a  into  two  parts,  so  that  if  one  of  them  be  di- 
vided hy  6,  and  the  other  by  c,  the  sum  of  the  quotients  shall 
be  d. 

48.  Separate  a  into  two  parts,  so  that  the  mth  part  of  one  shall 
exceed  the  nth  part  of  the  other  by  b, 

49.  What  number  is  that  whose  mth  part  exceeds  its  nth  part 
hy  pi 

50.  After  paying  away  —  and  -  of  my  money,  I  had  a  guin 
eaa  left.     How  many  guineas  had  I  at  first? 

51.  After  paying  away  the  —  and  the  -  parts  of  my  money, 

I  had  a  dollars  left.     How  much  money  had  I  at  first? 

52.  A  and  B  together  could  do  a  piece  of  work  in  a  days  ;  B 
could  do  it  alone  in  b  days ;  in  how  many  days  could  A  do  it 
alone  ? 

53  A  company  at  a  tavern  paid  a  shillings  each ;  but  if  there 
had  been  b  persons  less,  each  would  have  had  to  pay  c  shillings. 
How  many  persons  in  the  company  ? 


1 18  GENERALIZATION.  XXV  II 

54  A  gentleman  has  6  sons,  each  of  whom  is  a  years  older 
than  his  next  younger  brother,  and  the  eldest  is  b  times  as  old  aa 
ihe  youngest.     Required  their  ages. 

55.  A  person  borrowed  as  much  money  as  he  had  in  his  purse, 
and  then  spent  a  shillings ;  again  he  borrowed  as  much  as  he 
had  left  in  his  purse,  after  which  he  spent  a  shillings ;  he  bor- 
rowed and  spent,  in  the  same  manner,  a  third  and  fourth  time ; 
after  the  fourth  expenditure  he  had  nothing  left.  How  much 
money  had  he  at  first? 

56.  A  man  agreed  to  work  n  days,  with  this  condition,  that  he 
should  receive  a  shillings  for  every  day  he  worked,  but  should 
forfeit  h  shillings  for  every  day  he  was  idle.  At  the  end  of  the 
time  agreed  on,  he  received  a  balance  of  c  shillings.  How  many 
days  did  he  work,  and  how  many  was  he  idle  ? 

57.  A  gentleman  gave  some  beggars  a  cents  apiece  and  had  h 
cents  left ;  but  if  he  had  given  them  c  cents  apiece,  he  would 
have  been  obliged  to  borrow  d  cents  for  that  purpose.  How 
many  beggars  were  there? 

The  following  questions  may  be  solved  by  means  of  tfwo  un- 
known quantities. 

58.  Said  A  to  B,  the  sum  of  our  ages  is  a  years,  and  their  dif- 
ference is  b  years.     Required  their  ages,  A  being  the  elder. 

59.  One  pair  of  boots  and  a  pairs  of  shoes  cost  b  dollars;  and 
c  pairs  of  boots  and  one  pair  of  shoes  cost  d  dollars.  Required 
the  price  of  the  boots  and  shoes  a  pair. 

60.  There  are  two  numbers,  such  that  if  y  part  of  the  second 

be  added  to  the  first,  the  sum  will  be  a ;  and  if  -  part  of  the  firs^ 

c 

be  added  to  the  second,  the  sum  will  also  be  a.     Required  the 

wo  numbers. 

61.  What  fraction  is  that,  to  the  numerator  of  which  if  a  be 

added,  the  value  of  the  fraction  will  become  —  ;  but  if  a  be  ad 

n 

n 

ded  to  the  denominator,  the  value  of  the  fraction  will  be  —  T 

9 


XXVIII.  NEGATIVE    QUANTITIES,    ETC.  ^  119 

62.  What  fraction  is  that,  from  the  numerator  of  whith  if  a  be 
Bubtracted,  the  value  of  the  fraction  will  become  —  ;  but  if  a  be 
subtracted  from  the  denominator,  the  value  of  the  fraction  will 

P  n 

oecorae  -  s 
9 

63.  What  will  be  the  particular  answer  to  the  61st,  if  a=r  2 

—  z=  f ,  and  ~  =  -^;  and  what  will  be  the  particular  answer  lo 
the62d,ifa  =  3, -=T^,  and  ^  =  Wl 


SECTION    XXVlll. 

NEGATIVE    QUANTITIES    AND    THE    INTERPRETATION     OF     NEOATIVB 

RESULTS. 

Art.  T8.  It  may  happen,  in  consequence  of  some  absurdity 
or  inconsistency  in  the  conditions  of  a  problem,  that  we  obtain, 
for  a  result  or  answer  to  the  question,  a  quantity  affected  with 
the  sign  — .     Such  a  result  is  called  a  negative  solution. 

Negative  results  not  only  indicate  some  absurdity  or  incon^is 
ten,cy  in  the  conditions  of  a  question,  but  also  teach  us  how  to 
modify  the  question,  so  as  to  free  it  from  all  inconsistency. 

As  such  negative  quantities  frequently  occur,  we  shall  proceed 
to  show,  that,  when  isolated  or  standing  alone,  they  are  subject 
to  the  same  rules  as  when  connected  with  other  quantities. 

We  remark,  in  the  first  place,  that  negative  quantities  are  de- 
rived from  attempting  to  subtract  a  greater  quantity  from  a  les? 
The  greatest  quantity  that  can  be  taken  from  another,  is  that 
quantity  itself.  Thus,  7  is  the  greatest  number  that  can  be  sub- 
tracted from  7,  and  a  is  the  greatest  number  that  can  be  sul)- 
tractei'  from  a.  In  such  a  case  the  remainder  is  zero;  thu3^ 
7  —  7  =  0,  anda  — 0  =  0. 


i20  NEGATIVE    QUANTtTlES    AND    THE  XXVlll 

If  it  were  required  to  subtract  9  from  7,  we  represent  it  thus, 
7  —  9,  or  1  —  7  —  2;  this  being  reduced  becomes — 2.  Tha 
sign  —  before  the  2,  shows  that  there  were  2  out  of  the  9  units, 
which  could  not  be  actually  subtracted.  If  7  be  subtracted  from 
9,  the  remainder  is  the  same,  except  that  it  has  the  sign  -}-• 

In  like  manner,  if  we  subtract  b  from  a,  b  being  the  greater, 
the  remainder,  a  —  6,  will  be  negative;  but  if  we  subtract  a  from 
6,  the  remainder,  b  —  a,  will  be  the  same  as  before,  except  thai 
it  will  be  positive. 

Art.  79.  Suppose  it  were  required  to  add  b  —  c  to  a.  It  is 
evident,  that  we  are  to  add  to  a  the  quantity  6,  and  subtract  from 
the  sum  the  quantity  c,  and  the  result  is  a-\-b  —  c. 

Now,  as  the  reasoning  does  not  depend  at  all  upon  the  value 
of  6,  the  method  of  proceeding  must  be  the  same  when  6  =  0, 
which  reduces  the  expression  b  —  c  to  0  —  c  or  —  c,  and 
a-\-b  —  c  becomes  a-f-O  —  c  or  a  —  c;  that  is,  —  c  added  to 
a  gives  a  —  c,  which  accords  with  the  rule  already  given  for  ad- 
ding polynomials.     Hence, 

Adding  a  negative  quantity — c,  is  equivalent  to  subtracting 
an  equal  positive  quantity  -\-  c. 

Art.  80.  Since  b  —  6  =  0,  a-\-b  —  b  is  of  the  same  value  as 
a,  and  may  be  regarded  as  the  quantity  a  under  a  different  form. 
Now,  in  order  to  subtract  +  b  from  a,  it  is  sufficient  to  strike  it 
out  from  the  eixpression  a-\-b  —  6,  and  we  have  a  —  b\  or  if 
we  would  subtract  —  6,  strike  that  out,  and  we  have  a-\-b;  that 
is,  -\-b  subtracted  from  a  gives  a  — by  and  — b  subtracted  from 
a  gives  a-\-b,  which  accords  with  the  rule  already  given  for 
subtracting  polynomials.     Hence, 

Subtracting  a  negative  quantity  —  6,  25  equivalent  to  adding 
an  equal  positive  quantity  -\-  b. 

Art.  81.  With  regard  "to  multiplication,  in  Art.  30,  we  have 
alieady  seen  that  the  product  of  a  —  A  by  c  —  d  is 

ac  —  be  —  ad-^-bd. 


XX VIII.  INTERPRETATION    OF    NEGATIVE    RESULTS.  12i 

Now  it  is  manifest  that  the  sign  of  any  term  in  this  product 
IS  entirely  iiidependent  of  the  absolute  value  of  the  letters,  a,  6^ 
c,  and  d. 

Let  us  suppose  then,  in  the  first  place,  that  a  and  d  are  each 
equal  to  zero.  Upon  this  supposition,  the  quantities  to  be  mul- 
tiplied together  become  0  —  b  and  c  —  0,  or  — h  and  c;  and  the 
product  becomes  O.c  —  he  —  0.0-|-6.0,  or  —  be.  Hence, 
*- h  multiplied  by  -}- <^>  produces  —  he. 

Secondly,  suppose  6  and  c  each  equal  to  zero.  Then  the  quan 
tities  to  be  multiplied  together  become  a  and  — d\  and  the  pro- 
duct, ac  —  be  —  ad-\-bd^  is  reduced  to  —  ad.  Hence,  -{-  a 
multiplied  by  — d,  produces  — ad. 

Lastly,  let  the  value  of  each  of  the  letters  a  and  c  be  zero> 
We  then  have  to  multiply  — b  by  — d\  and  the  product,  ac  — 
he  —  ad-\-hd,  is  reduced  to  '■\- h  d.  Hence,  — 6  multiplied 
by  —  d,  produces  -\-hd. 

From  these  several  results  we  deduce  the  same  rule  for  the 
signs  in  multiplication,  as  that  given  in  Art.  31. 

Art.  SS.  Since  in  division,  the  product  of  the  divisor  and 
quotient  must  reproduce  the  dividend,  it  follows  from  the  prece- 
ding demonstration,  that  the  rule  for  the  signs  in  the  division  of 
isolated  quantities,  is  the  same  as  that  given  for  polynomials. 

We  conclude  then,  in  general,  that  the  four  fundamental  op- 
erations are  performed  upon  algebraic  quantities  when  isolated, 
according  to  the  same  rules,  in  respect  to  the  signs,  as  when  they 
constitute  terms  of  polynomials. 

Art.  83.  It  is  manifest  from  what  precedes,  that  addition  in 
algebra  does  not  always  imply  the  idea  of  augmentation;  for,  if 
we  add  — 6  to  a,  the  result  a  —  6  is  less  than  a  by  the  quan- 
t'ty  h. 

Nor  does  subtraction  in  algebra  always  imply  the  idea  of  dimi- 
nution ;  for,  if  we  subtract  —  h  from  a,  the  result  a-]- his  greater 
ihan  a  by  the  quantity  b. 

To  distinguish  these  results  from  those  of  addition  and  sub- 
traction in  arithmetic,  we  use  the  terms  algtbraie  sum  and  alge* 
11 


122  NEGATIVE    QUANTITIES    AND    THE  XXVIIl 

braic  difference.      Thus,  a  —  6  is  the  algebraic  sum  of  a  and 
—  b ;  and  a  +  6  is  the  algebraic  difference  between  a  and  —  b 

Art.  84:.  1.  If  a  rectangular  field  is  10  rods  long  and  7  rodji 
wide,  how  much  must  be  added  to  the  length,  in  order  that  the 
field  may  contain  49  square  rods  ? 

Suppose  X  rods  added  to  the  length ;  then 

10  -j-  2;  =  the  length  after  x  rods  are  added ;  hence, 

7(10  +  x)r=49,  or  70  +  7x  =  49. 

This  equation  gives  x  =  — 3  rods. 

Here  the  value  of  x  being  negative,  indicates  some  absurdity 
in  the  question. 

If  we  return  to  the  equation  70  -|-  7  x  =  49,  we  perceive  that 
the  absurdity  consists  in  supposing,  that  something  must  be 
arithmetically  added  to  70  to  make  it  equal  to  49. 

The  result,  2;  zi:  —  3,  shows  that  —  3  rods  must  be  algebraic- 
ally added  to  the  length,  that  is,  3  rods  must  be  subtracted 
from  it. 

Let  us  then  put  — x  instead  of  -|-x  in  the  original  equation, 
and  it  becomes 

7(10  — x)  =  49,  or  70  — 7x=:49. 

This  gives  x  r=  3  rods. 

The  question  therefore  should  have  been  as  follows : 

If  a  rectangular  field  is  10  rods  long  and  7  rods  wide,  how 
much  must  be  subtracted  from  the  length,  that  the  field  may  con- 
tain 49  square  rods  ? 

We  are  conducted  to  this  modification  in  the  question,  merely 
by  changing  the  sign  of  x  in  the  original  equation.  We  see, 
moreover,  that  both  equations  give  the  same  result,  except  with 
regard  to  the  sign. 

2.  If  a  field  is  9  rods  long  and  5  rods  wide,  how  much  must 
be  subtracted  from  the  length,  so  that  the  area  of  the  field  may 
be  65  square  rods  t 

Kx  =  the  number  of  rods  to  be  subtracted,  we  have 
5(9  — x)=65.  or  45— 5x=:65. 


XXVIII.        INTERPRETATION    OF    NEGATIVE    RESULTS.  123 

This  equation  gives  x  =r  —  4 ;  hence,  —  4  rods  *are  to  be  sub- 
tracted from  the  length,  that  is,  4  rods  are  to  be  added  to  i^ 

Indeed  the  equation  45 — 5x=i65  is  evidently  absurd,  since 
it  supposes  that  something  must  be  taken  from  45  to  make  it 
equal  to  65. 

Let  us  put  +  ^  instead  of  —  x  in  the  original  equation  ;  thiw 
equation  then  becomes 

5  (9  -|-  a;)  =  65,  which  gives  2=4  rods. 

We  see  therefore  that  the  inquiry  should  have  been,  how  much 
must  be  added  to  the  length. 

3.  A  father  whose  age  is  68  years,  has  a  son  aged  20;  iK 
how  many  years  will  the  son  be  one  fourth  as  old  as  his  father  t 

Suppose  X  =  the  number  of  years  ;  then 

68  4-x 
20  -f-  X  :=  — J—.     This  equation  gives  x  =  —  4. 

Changing  the  sign  of  x  in  the  original  equation,  we  have 

20  —  x  =  — J — ,  which  gives  x  =  4  years. 

The  question  therefore  should  have  been;  how  man/  yeara 
ago  was  the  son  one  fourth  as  old  as  his  father  ? 

4.  A  laborer  wrought  for  a  gentleman  7  days,  having  his  son 
with  him  4  days,  and  received  27  shillings ;  at  another  time  he 
wrought  9  days,  having  his  son  with  him  6  days,  and  received  33 
shillings.  What  were  the  daily  wages  of  the  laborer  and  his  sof 
lespectively  ? 

Let  X  =  the  daily  wages  of  the  man, 
and  y  =  the  daily  wages  of  the  boy. 
Hence,  7x-j-4y=:27, 
and         9x-t-6y  =  33. 
These  equations  give  x  =r  5,  and  i/=.  —  2. 
Changing  the  sign  of  y  in  the  original  equations,  we  have 
7x  — 4y  =  27, 
and  9x  — 6y  =  33. 
These  equations  give  x  =  5  shillings  and  y  =  2  shilhngs. 


124  NEGATIVE    QUANTITIES    AND    THE  XXVlll 

It  Appears  then  that  the  son  was  an  expense  to  his  father,  and 
that  the  inquiry  should  have  been :  how  much  did  the  laborer 
receive  per  day  for  himself,  and  how  much  did  he  pay  per  day 
for  his  son  ? 

5.  What  fraction  is  that,  to  the  numerator  of  which  if  1  be 
added,  the  value  of  the  fraction  will  be  f  j  but  if  1  be  added  to 
the  denominator,  the  value  will  be  f  ? 

Let     -  be  the  fraction. 

y 

Then  ^±J  =  I, 

y 

and     —^T—i  =  I-     These  equations  give  %=.  —  5, 

and  y  =  —  9. 

Here  the  values  of  x  and  y  are  both  negative.     Changing  the 

signs  of  X  and  y  in  the  original  equations,  we  have 

—  2:4- 1       ^         ,       — a;  ^       -r* 

=.  f,  and  -— -  =  #.      But  we  may 

—  y  — y+A 

change  the  signs  of  the  numerators  and  denominators  of  the  first 
members  without  altering  the  value  of  the  fractions;  we  then  have 

=  I,  and =  f . 

y  y— 1 

The  question  should,  therefore,  have  been  as  follows : 

What  fraction  is  that,  from  the  numerator  of  which  if  1  be 
subtracted,  the  value  will  be  | ;  but  if  1  be  subtracted  from  the 
denominator,  the  value  will  be  |? 

The  preceding  problems  render  it  manifest,  that  a  negative 
result  indicates  some  absurdity  in  the  conditions  of  the  question, 
and  show  us,  that  the  conditions  are  modified  so  as  to  remove  the 
absurdity,  by  rendering  subtractivCf  quantities  which  had  been 
previously  considered  as  additive,  or  by  rendering  additive,  quan- 
tities which  had  previously  been  considered  as  subtractive. 

We  see  moreover,  that,  in  order  to  ascertain  what  the  condi- 
tions should  have  been,  we  have  only  to  change,  in  the  original 
equations,  the  signs  of  those  (juantities  for  which  we  have  ob- 
tained negative  values,  ana  modify  the  question  accordingly. 


X^XVIII.         INTERPRETATION    OF    NEGATIVE    RESULTS.  lZ5 

Negative  quantities  are  sometimes  said  to  be  less  than  zero^ 
and,  in  an  algebraical  sense^  they  may  be  so  considered.  But 
strictly  speaking,  no  quantit}  can  be  less  than  zero.  When  we 
say,  for  example,  of  a  bankrupt,  that  he  is  worth  $5000  less  than 
nothing,  we  mean  simply,  that  he  owes  $5000  more  than  he  can 

pay- 
Negative  quantities  do  not,  in  reality,  differ  from  positive  quan 

tities,  and  are  merely  positive  quantities  taken  in  a  sense  differ 

ent  from  that  first  supposed. 

Let  the  learner  solve  the  following  questions,  and  show  how 

the  negative  results  are  to  be  interpreted. 

6.  What  number  is  that,  /^  of  which  exceeds  ^  of  it  by  5  ? 

7.  A  man,  when  he  was  married,  w^s  30  years  old,  and  his 
wife  20.  How  many  years  before  their  marriage  was  his  age  to 
hers  as  7  to  6  ? 

8.  What  fraction  is  that,  whose  value,  if  its  denominator  be 
diminished  by  2,  will  be  ^,  but  whose  value,  if  its  numerator  be 
dnninished  by  2,  will  be  ^^? 

9.  Find  two  numbers,  such  that  their  difference  shall  be  20, 
and  the  difference  between  6  times  the  greater  and  3*  times  the 
less  shall  be  96. 

10.  A  cistern  is  provided  with  two  stopcocks,  A  and  B,  through 
which  water  flows.  After  the  stopcock  A  had  been  open  5  min- 
utes, and  B  3  minutes,  there  were  found  24  gallons  in  the  cis- 
tern ;  but  if  A  had  been  open  7  minutes  and  B  5,  there  would 
have  been  32  gallons  in  the  cistern.  How  much  water  flows  into 
the  cistern  through  each  stopcock  in  a  minute? 

il.  Three  times  A's  money,  twice  B's,  and  four  times  C*a 
make  $13000;  four  times  A's,  three  times  B's,  and  twice  C'a 
make  $25000 ;  and  six  times  A's,  four  times  B's,  and  once  C's 
mane  $40000.     Required  the  estate  of  each. 
11* 


*26  DISCUSSION    OF    PROBLEMS.  XXIX 


SECTION    XXIX. 

DISCUSSIOX    OF    PROBLEMS. 

Art  85.  "When  a  question  has  been  resolved  generally,  that 
IS,  by  using  letters  to  represent  the  known  quantities,  we  some- 
times inquire  what  values  the  unknown  quantities  will  assume, 
in  consequence  of  particular  suppositions  with  regard  to  the 
known  quantities.  The  determination  of  these  values,  and  the 
interpretation  of  the  remarkable  results  which  we  may  obtain, 
constitute  what  is  called  the  discussion  of  the  problem. 

The  discussion  of  the  following  problem,  which  is  originally 
due  to  Clairaut,  presents  many  remarkable  circumstances. 

1st  case.  Two  couriers  set  out,  at  the  same  time,  from  two 
points,  A  and  B,  which  are  a  miles  asunder,  and  travel  towards 
each  other.  The  courier  from  A  goes  b  miles  per  hour,  and  the 
courier  from  B,  c  miles  per  hour.  At  what  distance  from  A 
and  B  will  they  meet? 


A  R  B 

Let  R  be  the  point  of  meeting.  Suppose*  z  =  the  distance 
from  A  to  R,  and  y  =  the  distance  from  B  to  R.  Then  we 
have 

(1)  z  +  y  =  a. 

As  the  courier  from  A  goes  b  miles  per  hour,  he  will  be  -j- 
hours  in  going  x  miles ;  in  like  manner,  the  courier  from  B  will 
be  -  hours  in  going  y  miles ;  and  since  they  are  equal  times  on 
the  road,  we  have 

(2)  r  =  -  •     Multiply  the  2d  equation  by  6, 

2  =  —  ;  substitute  this  value  of  x  in  the  let, 
c 


^XIX  DISCUSSION    OF    PROBLEMS.  127 


— —  -\-i/  =za'f  multiply  by  c, 
by-\-ci/=:ac,  or  (b-\-c)i/z=ac;  hence, 

Qi  C 

y  =  =  distance  of  the  point  of  meeting  from  B. 

h  y 

Substituting  the  value  of  y  in  the  equation  z  =  -^,  or  2  = 

c 

-  X  y,  we  have 
c 

b        ac  abc  ab  ' 

x=z~  .  =—. —  =  —77—, — ^  =  r- 1 — =  distance  of  the  point 
c     o-j-c       c(o-j-c)       b-\-c  ^ 

of  meeting  from  A. 

As  the  sign  —  does  not  occur  in  the  values  of  x  and  y,  these 
values  will  always  be  positive,  whatever  numbers  are  put  instead 
of  a,  b  and  c.  Indeed  it  is  evident,  that  since  the  couriers  travel 
towards  each  other,  they  must  necessarily  meet  between  A 
and  B. 

2rf  case.  Suppose  now  that  the  couriers,  setting  out  from  the 
points  A  and  B,  as  in  the  diagram  below,  proceed  both  in  the 
same  direction,  and  travel  towards  the  point  C,  at  the  same  rates 
as  before.  What  distance  will  each  travel  before  one  overtakes 
the  other  ? 

A  B  R  C 

Suppose  R  the  point  where  they  come  together.  Let  z  =  A  R, 
and  z^  =:  B  R.     Then, 

(1)  z  —  yz=.  a,  and 

(2)  -=^. 
^  '     be 

These  equations  being  solved,  give 

a 

ab  ;  ac 

-,  and  y  = 


b  —  c' ^—b—c 

Here  the  values  of  x  and  y  will  not  be  positive,  unless  b  ia 
greater  than  c ;  that  is,  unless  the  courier  from  A  travels  faster 
Ihan  the  courier  from  B.  ^  -r5=:?^— "i^- 


128  DISCUSSION    OF    PROBLEMS.  KXIX 


Suppose  b  =.  10,  and  c  =  8 ;  then  we  have 
10  a    _  10a_ 


10  —  8         2 

,  8a  8a       ^ 

But  if  we  suppose  that  b  is  less  than  c,  and  that  5  =:  8  and 
:=  10,  we  have 

8a  8a  .  , 

^=8^ro==¥  =  -^"'^^ 

10a  lOa  ,    . 

-5  a. 


^      8—10       —2 

Here  the  values  of  x  and  y  are  both  negative,  and  show  that 
there  is  some  inconsistency  in  the  question  ;  and  indeed  it  is  ab- 
surd to  suppose,  that  the  courier  from  A  can  overtake  the  courier 
from  B,  both  travelling  towards  C,  unless  the  former  travel  faster 
than  the  latter. 

In  order  to  see  how  the  question  is  to  be  modified,  let  us 
change  the  signs  of  x  and  i/  in  the  original  equations. 

We  then  have 

y  —  X  zr:  a,  and 

-7—  =.  — -.     The  last  equation,  by  a  change  of 

the  signs,  becomes 

b  ~   c 

It  is  evident  that  the  2d  equation  will  remain  the  same  as  be- 
fore,  because  it  merely  expresses  the  equality  of  the  times 

Tue  equation  y  —  xz=a  shows  that  y  is  greater  than  x,  or 
thit  the  point  where  they  come  together,  is  further  from  B  than 
It  is  from  A ;  and  since  this  point  cannot  be  between  A  and  B, 
it  must  be  on  the  other  side  of  A  with  respect  to  B,  as  at  R'  in 
Ihe  subjoined  diagram. 


C  R'  A  B  R  C 

When,  therefore,  6  is  less  than  c,  the  question  should  be  aa 
follows : 


XXIX  DISCUSSION    OF    PROBLEMS.  129 

Two  couriers  set  out  from  the  points  A  and  B,  a  miles  distanl 
from  each  other,  and  travel  towards  C ;  the  courier  from  A  goes 
b  miles  per  hour,  and  the  courier  from  B,  c  miles  per  hour  ;  how 
far  will  each  travel  before  the  courier  from  B  overtakes  the  one 
from  A  ? 

[n  this  case,  the  equations, 

y  —  2  =  «,  and  -  =  -,  will  give 

ah  .  ac 

X  = -,  and  ij  = . 

c  —  6  c  —  6 

If  6  =  8  and  c  =  10,  we  have 

8a     ■  ^  10  a 

We  see,  in  this  question,  that  a  change  of  sign  indicates  a 
change  in  direction.  Numerous  instances  of  similar  indications 
occur  in  the  application  of  algebra  to  geometry. 

^d  case.     Resuming  the  formulae, 

ab          .             ac      .  .  . 

X  =: ,  and  i/  = ,  let   us   suppose   b  =z  c  ;    then 

b  —  c  being  0,  we  have 

ab        -  ac 

In  order  to  interpret  these  results,  let  us  go  back  to  the  orig- 
inal equations,  x  —  y  z=.  a^  and  -  =  -.     By  putting  b  instead 

of  e  in  the  second  equation,  it  becomes  -  in  -,    which    gives 

i:=.y\  and  substituting  x  for  y  in  the  first  equation,  we  have 
X  —  xz=.a,  or  0  :=z  a. 
This  result  is  manifestly  absurd,  since  we  have  a  known  quan- 
tity equal  to  zero ;  and  it  is  evident,  that  since  the  couriers  travel 
equally  fast,  it  is  impossible  one  should  overtake  the  other 

TTl 

We  regard  therefore  — ,  or  any  similar  expression,  as  a  syrri' 
bol  of  impossibility  ;^a.nd  when  a  question  gives  Ozna  (a  being 


130  DISCUSSION    OF    PROBLEMS.  XXIX 

any  known  |uantity  different  from  zero),  or  when  the  unknown 
quantity  is  found  =  -,  the  question  is  to  be  considered  as  im- 
possible. 

There  is,  however,  another  signification  of  such  an  expression 

m       .....    . 

as  —  ,  which  It  IS  important  to  notice. 

Let  us  take  the  expressions  for  x  and  y,  viz  :  2  = ,  and 

o '     c 


10  a  10  a       ,.^ 


Making  6  =  10  and   1      ""10  — 99         '1 

c  =z 9-9,  we  have       \  99 a  99 a      ^ 

_      10 a     _  lOf  —  innn 
Making  6  =  10  and   1  ""—lO  — 9  99~    01  —'""""' 

c  =  9-99,  we  have      \  999  a         999 a      ,^^ 

'»"       =15,f=10000«; 


Making  6  =  10  and   1      —  10  —  9-999  —  -001 

c  =z 9-999,  we  have    i  9999 a          9999  a      „-. 

( ^-ro'=9^999--^oor=^^^ 

We  here  perceive  that  the  value  of  the  fraction  increases  in  pro-' 
portion  as  the  denominator  is  diminished ;  if  then  the  denomina- 
tor be  less  than  any  assignable  quantity,  that  is  0,  the  value  of 
the  fraction  will  be  greater  than  any  assignable  quantity,  or  infi»« 
nitely  great.  Hence  mathematicians  consider  a  fraction,  whose 
lumerator  is  a  definite  quantity,  and  whose  denominator  is  zero, 

as  a  symbol  of  infinity.     Thus,  Tit  Tit      7"    ,  are  Symbols  of  in* 
finity. 

Remark.  In  the  problems  of  geometry,  solved  by  the  aid  of 
algebra,  there  are  many  instances,  in  which  an  in/inite  quantity 
Mislead  of  denoting  an  absurdity,  is  the  true  Result  sought. 


XX^X  DISCUSSION    OF    PROBLEMS.  131 


If  a  definite  quantity  be  divided  by  an  infinite  or  impossible 
quantity,  the  quotient  will  be  zero.     Thus  a  divided  by  -   gi'ei 

m         m 

4tth  case.     Suppose  now,  that,  in  the  formulae  for  2  and  y, 

6  =  c  and  a  =  0. 

The  distance  between  the  points  A  and  B  being  nothing,  thesa 

points  must  be  coincident,  as  in  the  following  diagram, 

A 

B 

and  the  formulae  for  x  and  y  become 

0.6       0        ^  O.c       0  n      A     u      . 

x=r  —  =  -,  and  y  =  —  =  -;  or  x  .  0  =  0,  that  is, 

0  =  0,  and  y  .  0  =  0,  that  is,  0  =  0. 

Now,  as  the  couriers  set  out  from  the  same  point,  and  travel 
equally  fast  and  in  the  same  direction,  they  cannot  be  said  to 
come  together  at  any  particular  point,  since  they  are  constantly 
together  throughout  the  whole  route. 

But  in  order  to  see  the  general  import  of  the  expression  ^,  let 

us  return  to  the  original  equations,  x  —  y^=.a^  and  -  =  -., 

Putting  0  instead  of  a,  and  h  instead  of  c,  we  have  x — y  =  0, 

,  «        y 
and-  =  -. 

The  first  equation  gives  x  =  y,  and  this  value  of  %  being  sub- 

y     V 

Btituted  in  the  second,  gives  -r=z^. 

This  last  equation,  in  which  the  two  members  are  precisely 
alike,  is  called  an  identical  equation,  and  is  verified  by  putting 
any  quantity  whatever  instead  of  y.  The  value  cf  t  therefore 
cannot  be  determined  from  this  equation. 

Moreover  the  equation  -  =  | ,  gives   x  =  y,    an<?  therefore 

expresses  nothing  more  than  the  first. 


rJ2  DISCCSSiON    OF    PROBLEMS.  XXIX 

Hence  the  expression  ^  is  a  symbol  of  an  indeterminate  quan* 
tity ;  and  when  a  problem  results  in  giving  0  =  0,  or  when  the 
unknown  quantity  is  found  ==  ^,  the  question  is  to  be  regarded 
as  indeterminate. 

There  are  however  some  precautions  to  be  used,  before  we 
decide  that  a  quantity  is  indeterminate. 

a3 ^3 

Thus, r-,  when  a  ==  i,  becomes  % ;  but  i  he  numerator 

a  —  0 

and  denominator  both  being  divisible  by  a  —  6,  if  the  fraction  is 
reduced,  it  becomes  a^-^-ah-^-hl^  or  Sa^,  which  is  a  deter- 
minate quantity. 

When,  therefore,  we  arrive  at  a  result  :=  %,  before  we  pro- 
nounce it  indeterminate,  we  must  see  whether  the  fraction  which 
represents  this  result,  has  not  a  factor  common  to  its  numerator 
and  denominator,  which  being  struck  out,  will  render  the  quan- 
tity definite. 

Let  the  learner  solve  the  following  problems  and  interpret  the 
results. 

1.  A  boy  being  asked  how  much  money  he  had,  replied,  that 
^  and  Y^2  of  his  money,  increased  by  40  cents,  would  be  equal  to 
I  of  his  money,  increased  by  49  cents.  How  many  cents  had 
he? 

2.  Four  men,  A,  B,  C  and  D,  talking  of  their  ages,  it  was 
found  that  B  was  10  years  older  than  A  and  10  years  younger 
than  C,  and  that  D  was  34  years  younger  than  A ;  moreover, 
that  ^  of  B's  age,  f  of  C's,  and  ^  of  D's  would  be  equal  to  twice 
A's  diminished  by  10  years.     Required  the  age  of  each. 

3.  How  many  ducks  have  you  killed  to-day,  said  a  farmer  to  a 
sportsman ;  the  latter  replied,  one  half  of  the  number  I  have 
killed  to-day,  exceeds  ^  of  what  I  killed  yesterday  by  5  ;  and  the 
number  I  killed  yesterday,  is  5  less  than  once  and  a  half  the 
number  I  have  killed  to-day.  Required  the  number  he  killed 
each  day. 


2LX.X       EXTRACTION  OF  THE  SECOND  ROOTS  OF  NUMBERS.  1311 


SECTIjON   XXX. 

EXTRACTION    OF    THE    SECOND    ROOTS    OF    NUMBERS. 

Art  86.  What  number  is  that,  which,  being  multiplied  by  7 
limes  itself,  gives  a  product  equal  to  448 1 

Let  X  represent  the  number ;  then  x  .7  x  or  7  x^z=z  448. 

This  is  called  an  equation  of  the  second  degree,  because  it  con- 
tains the  second  power  of  the  unknown  quantity.  Such  an  equa- 
tion is  also  sortietimes  called  a  pure  quadratic  equation. 

In  order  to  solve  this  equation,  we  first  divide  by  7,  and  have 
x^  =. 64,  or  X  .x=. 64. 

Hence,  x  must  be  a  number,  which  multiplied  by  itself,  will 
give  64 ;  and  we  know  that  8  .  8  =  64 ;  therefore  2  =  8. 

The  first  power  of  a  quantity,  in  reference  to  the  second,  is 
called  the  root,  and  finding  the  first  power  when  the  second  is 
given,  is  called  extracting  the  second  root.  The  second  root  of 
a  quantity  then,  is  such  as  being  multiplied  by  itself,' will  pro- 
duce the  given  quantity. 

The  second  powers  of  the  first  nine  figures,  are  as  follows. 

(  1, 2,  3,    4,    5,    6,    7,    8,    9.     Roots. 
(  1,  4,  9,  16,  25,  36,  49,  64,  81.     Powers. 

We  perceive  from  this  table,  that  when  a  number  contains 
only  one  figure,  its  second  power  cannot  contain  more  than  two 
figures.  The  least  number  containing  two  figures  is  10,  the  sec- 
ond power  of  which  is  100,  consisting  of  three  figures. 

In  order  to  find  a  rule  for  extracting  the  roots  of  numbers  con- 
taining more  than  two  figures,  let  us  see  how  a  second  power  is 
formed  from  its  root. 

The  second  power  of  a-^-b  is  a^  -{-  2  a  6  -f-  6^.  Suppose  a 
=  20  or  2  tens,  ani  6  =  5 ;  then  a^  =  400,  2a6  =  2.20.5  = 
200,  and  6^  —  25  hence  a^  _|_  2  a  6  -f  52  _.  400  _(-  200  +  25 
-=:  625. 

12 


134  EXTRACTION    OF    THE  XX^ 

TVhen,  therefore,  a  number  contains  tens  an^  units ,  its  second 
power  win  contain  the  second  power  of  the  tens,  plus  tioice  the 
product  of  the  tens  by  the  units, ^plus  the  second  power  of  the 
units. 

Now  let  us  reverse  the  process,  and  see  by  what  means  the 
root  could  be  found  from  the  power. 
Operation. 

6'25(25.     Root.  25  .  25  =  625. 

4 

22'5(4     Divisor. 

Since  the  second  power  of  the  tens  of  the  root  can  contain  no 
significant  figure  below  hundreds,  it  must  be  found  in  the  6,  that 
is,  6  (hundreds) ;  we  therefore  separate  the  last  two  figures  from 
the  6  by  an  accent,  placed  over  the  top.  Now,  because  the  sec- 
ond power  of  3  (tens)  is  9  (hundreds),  and  that  of  2  (tens)  is  4 
(hundreds),  the  latter  is  the  greatest  second  power  of  tens  con- 
tained in  6  (hundreds),  and  the  root  is  2  (tens).  We  place  2  at 
the  right  of  the  proposed  number,  separating  it  by  a  line,  as  is 
done  with  the  quotient  in  division,  and  subtract  the  second 
power,  4  (hundreds)  or  a^,  from  6  (hundreds). 

To  the  right  of  the  remainder  2,  we  bring  down  the  two  figures 
jbut  off,  and  have  225.  This  number  corresponds  to  2  a  6  -|-  6^  j 
that  is,  it  contains  twice  the  product  of  the  tens  of  the  root  by 
the  units,  plus  the  second  power  of  the  units.  If  it  contained 
2  a  6  only,  or  twice  the  product  of  the  tens  by  the  units,  we 
should  obtain  the  units  exactly  by  dividing  by  2  a,  or  twice  the 
tens.  As  it  is,  if  we  divide  by  twice  the  tens,  disregarding  the 
remainder,  we  shall  obtain  the  units  exactly,  or  a  number  a  little 
too  great. 

But  since  twice  the  tens  multiplied  by  the  units,  cannot  have 
any  significant  figure  below  tens,  if  we  take  4  merely  as  the  di- 
visor, we  must  reject  the  right  hand  figure,  5,  of  the  dividend. 
Or,  m  other  words,  since  the  divisor  is  ten  times  too  small,  if  we 
make  the  dividend  ten  times  too  small,  the  quotient  will  not  be 
ftffected  by  this  change.     The  divisor  4  is  contained  in  22  five 


XXX.  SECOND  ROOTS  OF  NUMBERS.  135 

times.  Putting  5  at  the  rig?  t  of  the  2  in  the  root,  we  have  25^ 
which  raised  to  the  second  power  gives  625.  Hence  25  is  the 
root  sought. 

We  shall  now  explain  how  the  correctness  of  any  figure  in 
the  root  may  be  ascerta  ned,  without  raising  the  whole  to  the 
second  power. 

Let  it  be  required  to  extract  the  root  of  1521. 
•  Operation. 

15'21(39.     Root. 
9 

62'1(69.     Divisor. 
621 
0. 

Reasoning  as  before,  we  find  the  greatest  second  power  of 
tens  contained  in  15  (hundreds),  to  be  9  (hundreds),  the  root 
of  which  is  3  (tens) ;  putting  3  as  the  first  figure  of -the  root,  and 
subtracting  its  second  power  from  15,  we  bring  down  the  next 
two  figures,  and  have  for  a  dividend  621.  This  corresponds  to 
2  a  6  +  ^^>  which  is  the  same  as  6  (2  a  +  6).  Dividing  62  by  6, 
twice  the  tens,  we  have  for  a  quotient  10 ;  but  as  the  unit  figure 
cannot  exceed  9,  we  put  9  in  the  root  at  the  right  of  the  3,  and 
we  have  39  for  the  entire  root. 

In  order  to  determine  whether  9  is  the  proper  unit  figure  of 
the  root,  we  observe  that  the  divisor  6  (tens)  corresponds  to  2  a, 
and  9  is  the  figure  which  we  have  found  for  h ;  hence,  60  -f-  9 
or  69  corresponds  \.o2a-\-h\  therefore  we  place  9  at  the  right 
of  the  divisor  and  multiply  69  by  9;  the  product  621  answers  to 
6  (2  a  -|-  6) ;  this  subtracted  from  the  dividend  leaves  nothing. 
Therefore  the  true  root  is  39. 

Let  the  learner  extract  the  roots  of  the  following  numbers  bf 
the  process  last  explained. 
'     1.     784.  4.     841. 

2.  2809.  5.     1296 

3.  6084  6.     8649 
7    What      the  second  root  of  127449? 


136  EXTRACTION    OF    THE  XXX 

The  second  {owers  of  10,  100,  1000  are  respective.y  100 
10000,  1000000;  hence  the  second  povve  of  any  whole  numbef 
between  10  and  100,  that  is,  consisting  of  two  figures,  will  be 
between  100  and  10000,  that  is,  it  will  contain  three  or  four  fig- 
ures; also,  the  second  power  of  a  number  consisting  of  three 
figures,  will  contain  five  or  six  figures.  We  canj  therefore,  as- 
certain the  number  of  figures  in  the  root  of  any  proposed  num- 
ber, by  beginning  at  the  right,  and  separating 'it  into  parts  or 
periods  of  two  figures  each.  The  left  hand  period  may  consist 
of  one  or  two  figures.  There  will  be  as  many  figures  in  the 
root,  as  there  are  periods  in  the  power. 

Separating  127449  into  periods,  we  see  that  the  root  must 
contain  three  figures,  or  hundreds,  tens,  and  units. 

Let  a  represent  the  hundreds  of  the  root,  b  the  tens,  and  c  the 
units ;  then  a-\-h-\-c^  regard  being  paid  to  the  rank  of  the  fig- 
ures, will  represent  the  root. 

The  second  power  ofa-|-6-f~^  's  a^-\-^ak-\-h'^ -\-^cic-\- 
2hc-\-c^,  or  a2  +  2a6  +  62-|-2{«  +  6)c  +  c2.  Hence,  the 
second  power  contains  the  second  power  of  the  hundreds,  plus 
twice  the  product  of  the  hundreds  by  the  tens,  plus  the  second 
power  of  the  tens,  plus  twice  the  sum  of  the  hundreds  and  tens 
multiplied  by  the  units,  plus  the  second  power  of  the  units.  We 
proceed  now  to  extract  the  root. 

Operation. 
1274^49(357.     Root. 

_9 =  «2. 

37'4(65        i=::2«  +  6. 
325  =:(2«  +  6)6. 

494'9(707  =  2(a  +  6)  +  c. 
4949    •      =:[2(«  +  6)  +  c]c. 
0. 
We  first  seek  the  second  power  of  the  hundreds  of  the  root, 
which  must  be  foui?d  in  the  12,  (120000) ;    the  greatest  second 
power  in  this  part   i  9,  (90000),  the  root  of  which  is  3,  (300). 


XXX  SECOND    ROOTS    OF    NUMBERS.  137 

Putting  3  as  the  first  figure  of  ^he  root,  and  subtrac  in^  its 
second  power  from,  12,  we  bring  down  the  next  period  at  the. 
right  of  the  remainder.     Wp  now  consider  374  as  a  dividend. 

This  dividend  contains  2ab-^  b^,  or  twice  the  product  of  the 
hundreds  by  the  tens,  plus  the  second  power  of  the  tens,  togethei 
with  the  hundreds  arising  from  multiplying  twice  the  sum  of  the 
hundreds  and  tens  by  the  units. 

It  is  now  our  object  to  find  6,  or  the  tens  of  the  root ;  and  for 
this  purpose,  we  divide  by  2  a  or  twice  the  hundreds.  But  as 
the  product  of  twice  the  hundreds  by  the  tens,  can  have  no  sig- 
nificant figure  below  the  fourth  place,  in  dividing  we  reject  the 
right  hand  figure  of  the  dividend,  separating  it  by  an  accent. 

We  double  the  hundreds,  and  obtain  6  for  a  divisor,  which  is 
contained  in  37  six  times. 

But  if  we  put  6  at  the  right  of  the  divisor  and  multiply  66  by 
6,  we  obtain  a  product  greatef  than  374.  We  next  try  5,  which 
we  place  in  the  root  and  also  at  the  right  of  the  divisor,  and  we 
have  65  corresponding  to  2  a  -\-b;  this  multiplied  by  5  gives 
325,  corresponding  to  (2  a  -f-  b)  b. 

We  now  subtract  325  from  the  dividend,  to  4;he  remainder 
annex  the  figures  of  the  last  period,  and  obtain  for  a  new  divi- 
dend 4949. 

This  dividend  contains  2  («  +  6)  c  +  c^  —  [2  (a  +  6)  +  c]  c, 
or  twice  the  sum  of  the  hundreds  and  tens  multiplied  by  the 
units,  plus  the  second  power  of  the  units.  To  obtain  the  units, 
therefore,  we  must  divide  by  twice  the  hundreds  and  tens  already 
found. 

But  as  hundreds  and  tens  multiplied  by  units,  can  have  no 
significant  figure  below  tens,  we  reject  the  right  hand  figure  of 
the  dividend,  separating  it  by  an  accent.  Double  the  hundreds 
and  tens  makes  70,  (700),  =  2  («-|- 6),  which  is  contained  in 
494,  (4940),  seven  times. 

We  then  put  7,  which  corresponds  to  c,  in  the  root,  and  also 
at  the  right  of  the  divisor,  and  we  have  707  =  2  («-|- 6) -{- c; 
this  multiplied  by  7  gives  4949  zzi  [2  {a -\-  b) -\-  c]  c,  which  sub- 
12* 


138  EXTRACTION    OF    THE  XXX 

tracted  from  the  dividend,  heaves  no  remainder.     Therefore  357 
IS  the  root  sought. 

If  the  root  contains  more  than  three  figures,  representing  the 
figure  of  the  highest  rank  by  a,  the  next  by  h,  &c.,  we  have 
for  the  second  power  a^  -\- '^  a  h  -\-  h'^  -\-  '^  [a  ~\- h)  c  -\-  c^  -\- 
2(a  +  6  +  c)^+rf2_|_2(a+i  +  c  +  (/)e  +  c2,  &c.,  ona2^ 
(2a  +  6)6+  [2(a  +  fe)  +  c]  c+  [2(a  +  6  +  c)  +  ^]  rf  + 
\jl  {a -\- b  -\-  c -\- d) -\- e]  e  &lc.  ;  the  first  form  of  which  shows, 
that,  after  the  first  figure  has  been  found,  each  of  the  successive 
figures  is  obtained  by  dividing  by  twice  the  whole  root  already 
found;  and  the  second  form  shows,  that,  in  each  case,  the  quo- 
tient is  to  be  placed  at  the  right  of  the  divisor,  and  that  the  divi- 
sor thus  increased,  is  to  be  multiplied  by  the  quotient. 

Moreover,  from  a  consideration  of  the  rank  of  the  figures,  it  is 
plain,  that  twice  the  root  already  found,  multiplied  by  the  next 
lower  figure,  can  produce  no  significant  figure  below  the  second 
from  the  right  in  each  dividend. 

a  What  is  the  second  root  of  1024832169? 
Operation. 
1024832169(32013.     Root. 

12'4(62 
124  • 

8321(6401 

6401 

192069(64023 

192069 


0. 
^neraimg  in  this  question  as  in  the  preceding  ones,  we  find 
that  the  second  divisor  64,  is  not  contained  in  the  dividend  83,. 
the  right  hand  figure  being  rejected,  which  shows  that  there  are 
no  hi»ndreds  in  the  root  sought ;  in  this  case,  we  place  a  zero  in 
the  root,  also  at  the  right  of  the  divisor,  and  bring  down  the  nexf 
two  fip'Tres  to  form  a  dividf  id. 


XXX  SECOND    ROOTS    OF    NUMBERS.  l39 

We  may  observe,  that,  if  the  last  figure  of  the  preceding  aivi- 
sor  be  doubled,  the  root,  so  far  as  it  is  ascertained,  will  be 
doubled ;  for  that  divisor  contains  twice  this  root,  with  the  ex- 
ception of  the  figure  last  found. 

Art.  87.  From  the  preceding  analysis  we  derive  the  follow- 
ing 

RULE    FOR    EXTRACTING    THE    SECOND    ROOTS    OF    NUMBERS. 

1 .  Begin  at  the  right,  and,  by  means  of  accents,  separate  the 
number  into  periods  of  two  figures  each.  The  left  hand  period 
may  contain  one  or  two  figures. 

2.  Find  the  greatest  second  power  in  the  left  hajid  period, 
place  its  root  at  the  right  of  the  proposed  number,  separating  it 
by  a  line,  and  subtract  the  second  power  from  the  left  hand 
period. 

3.  To  the  right  of  the  remainder  bring  down  the  next  period 
to  form  a  dividend.  Double  the  root  already  found  for  a  divisor. 
Seek  how  many  times  the  divisor  is  contained  in  the  dividend,  re- 
jecting  the  right  hand  figure.  Place  the  quotient  in  the  root,  at 
the  right  of  the  figure  previously  found,  and  also  at  the  right  of 
the  divisor.  Multiply  the  divisor  thus  increased  by  the  last 
figure  of  the  root,  and  subtract  the  product  from  the  whole  divi- 
dend. 

4.  Bring  down  to  the  right  of  the  remainder  the  next  period, 
to  form  a  new  dividend.  Double  the  root  already  found  for 
a  divisor,  and  proceed  as  before  to  find  the  third  figure  of  the 
root. 

Repeat  this  process  until  all  the  periods  have  been  brought 
down. 

Remark.  If  the  dividend  will  not  contain  the  divisor,  the 
right  hand  figure  of  the  former  being  rejected,  place  a  zero  in 
the  root,  also  at  the  right  of  the  divisor,  and  bring  down  the  next 
tteriod. 

Extract  tfte  roots  of  the  following  numbers. 


140         EXTRACTION  OF  THE  SECOND  ROOT&   OF  NUMBERS.       XXX 

1.  1369.  ,  7.     36100 

2.  2401.  8.     1100401 
3     361.                                       9.     1432809. 

4.  123201  10      151905625 

5.  502681.  11.     901260441. 

6.  11881.  12.     2530995481. 

Art.  88.  There  are  comparatively  but  few  numbers  which 
are  exact  second  powers ;  and  the  roots  of  such  as  are  not  per* 
feet  powers,  cannot  be  obtained  exactly  either  in  whole  numbers 
or  fractions.  For  example,  the  root  of  42  is  between  6  and  7 ; 
but  no  number  can  be  found,  which,  multiplied  by  itself,  will 
produce  exactly  42.  We  shall  however  see  hereafter,  that  the 
root  of  any  positive  number  may  be  approximated  to  any  degree 
of  exactness. 

Since  the  roots  of  numbers,  which  are  not  perfect  powers,  can- 
not be  obtained  exactly,  either  in  whole  or  fractional  numbers, 
they  are  said  to  be  irrational^  or  incommensurable ;  that  is,  these 
roots  and  unity  have  no  common  divisor.  Roots  of  other  de- 
grees, besides  the  second,  are  also  called  irrational ^  when  they 
cannot  be  exactly  obtained. 

The  second  root  of  a  quantity,  whether  that  root  can  be  found 
exactly  or  not,  is  indicated  either  by  the  exponent  ^,  or  by  this 
character  ^,  called  the  radical  sign.  Thus,  (25)^^  or  i/25 
=r  5 ;   and  (3)^  or  ^3    means  the  second  root  of  3. 

But  the  second  root  of  a  negative  quantity  cannot  be  obtained, 
even  by  approximation,  since  there  is  no  number,  which,  multi- 
plied by  itself,  can  give  a  negative  quantity.  The  second  roots, 
therefore^  of  negative  quantities  are  called  imaginary,  in  opposi- 
tion to  those  of  positive  quantities,  which  are  rcaZ,  although  they 

cannot  be  exactly  obtained.  Thus,  ( — 16)^  or  ^ — 16  is  im- 
aginary. These  imaginary  quantities,  except  in  some  of  the 
higher  branches  of  analysis,  indicate  absolute  absurdity  in  the 
questions  from  wHch  they  arise. 


J 


fi)' 


XXXI.  SECOND    ROOTS    OF    FRACTIONS,    ETC.  141 


SECTION    XXXI. 

•  BCOND    ROOTS     OF    FRACTIONS  —  AND    THE    EXTRACTION    OP    SECOND 
ROOTS    BY    APPROXIMATION. 

Art.  89.    The  second  power  of  a  fraction  is  found  by  raising 
both  numerator  and  denominator  to  the  second  power ;  for  this 
ia  equivalent  to  multiplying  the  fraction  by  itself.     Thus, 
^        a     a       a^ 
h'  b~W 

Hence,  the  second  root  of  a  fraction  is  found  by  extracting 
the  root  of  the  numerator  and  that  of  the  denominator.     Thus, 

the  root  of  ^f  is  ^,  and  that  of  -^  is  -. 

Let  the  roots  of  the  following  fractions  be  found. 
1-     a-  4.     iff. 

2.  «.  5.     /^V- 

3.  T%V-  6.     T^V- 

Art.  90.  If,  however,  either  the  numerator  or  denominator  is 
not  a  perfect  second  power,  the  root  of  the  fraction  can  be  ob- 
tained by  approximation  only.  Thus,  the  root  of  ^f  is  between 
^  and  ^.     It  is  nearer  to  f . 

The  denominator  of  a  fraction,  however,  may  always  be  ren- 
dered a  perfect  second  power,  by  multiplying  both  numerator 
and  denominator  by  the  denominator,  which  does  not  alter  the 
value  of  the  fraction.  For  example,  f  =  f ^,  the  approximate 
root  of  which  is  f  — .  By  this  mode,  the  root  has  the  same  de- 
nominator as  the  given  fraction. 

Re7nark.  The  sign  -|-,  placed  after  an  approximate  root,  sig- 
nifies that  it  is  less,  and  the  sign  — ,  that  it  is  greater  than  the 
true  one. 

When  a  greater  degree  of  exactness  is  requisite,  we  may,  aftei 
having  multiplied  both  terms  of  the  fraction  by  its  denominate f, 
multiply  both  terms  of  the  result  by  any  second  power. 


142  APPROXIMATE    SECOND    ROOTS    OF    NUMBERS.  XXXI 

3       21        144.21       3024      , 
^hus,  ^  =  49  =  1^14749  or  ^^,  the  approximate   -oot  of 

which  IS  —  — . 

o4 

Let  the  learner  find  the  roots  of  the  following  fractions,  in  the 
denomination  of  their  respective  denominators. 

1.  J.  4.     ^. 

2.  ^,  5.     f^. 

3.  A.  6.     U. 

Art  Ol.  The  root  of  a  whole  number  may  be  approximated 
in  the  same  way,  by  converting  it  into  a  fraction,  having  a  sec- 
ond power  for  its  denominator.  If,  for  example,  we  would  find 
the  root  of  5,  exact  to  12ths,  we  change  5  to  the  fraction  ^^, 
the  approximate  root  of  which  is  f|-  — . 

But  it  is  most  convenient  to  change  the  number  into  a  frac- 
tion, having  the  second  power  of  10,  100,  or  1000,  &,c.,  for  a  de- 
nominator ;  that  is,  convert  the  number  into  lOOths,  lOOOOths, 
or  lOOOOOOths,  &c.,  and  the  root  will  be  found  in  decimals. 
'  Thus,  5  =  U^  =  mU  =  mUU ;  that  is,  5  =  5-00  = 
5-0000  z=:  500000Q  ;  the  approximate  root  of  the  first  is 
f  f  +  =  2-2  +,  of  the  second  f  f  ^  +  =  223  +,  and  of  the  third 
fHf  +  =  2-236+. 

In  the  example  just  given,  we  perceive  that  twice  as  many 
/eios  are  annexed  to  the  number,  as  we  wish  to  have  decimals 
in  the  root.  Indeed,  it  is  plain,  that  there  must  be  half  as  many 
decimals  in  the  root  as  there  are  in  the  power,  because  the  sec- 
ond power  of  lOths  produces  lOOths,  the  second  power  of  lOOtha 
produces  lOOOOths,  &c. 

Moreover,  we  nee<l  not  add  all  the  zeros  at  once,  but  may 
annex  two  to  each  remainder,  in  the  same  manner  as  we  bring 
di»wn  the  figures  of  successive  periods. 

As  an  example,  let  us  extract  the  second  root  of  3. 


XX  Xi.  APPROXIMATE    SECOND    ROOTS    OF    NUMBERS  143 

Operation. 
3(1-732  + 

1 

20'0(27 
189 
1100(343 
1029 
710^0(3462 
6924 


176. 

The  operation  might  be  continued  to  any  extent. 

The  process  will  be  the  same  for  any  number  containing  deci» 
mals;  and  any  fraction  may  be  converted  into  decimals,  and  the 
root  may  be  extracted  in  the  same  way,  care  being  taken  to  make 
the  number  of  decimals  even. 

It  is  best,  when  the  number  contains  decimals,  to  begin  at  the 
decimal  point,  and  separate  the  decimals  into  periods  by  proceed- 
ing towards  the  right,  and  the  whole  numbers  by  proceeding 
towards  the  left. 

Art.  03.  In  approximating  a  second  root,  we  may  sometimes 
be  in  doubt,  whether  the  last  figure  found  is  so  great  as  it  should 
be.     This  may  be  determined  in  the  following  manner. 

The  second  power  of  a  is  a^,  and  that  ofa  +  l.is  «2-|-2a 
-|-  1.  Now  the  root  of  a^  -j-  2  a  -{~  1  ^^  ^  ■+-  1 5  but  if  we  should 
call  a  its  root,  and  subtract  the  second  power  of  a,  there  would 
remain  2  «  + 1.  Hence,  when  the  root  admits  of  being  increased 
by  I,  the  remainder  will  contain  at  least  twice  the  root  already 
found  plus  1,  the  local  value  of  the  figures  being  disregarded. 

Thus,  if,  in  the  last  example,  we  had  obtained  1*731  instead 
of  1-732  for  the  root,  the  remainder  would  have  been  3639,  which 
exceeds  twice  1731  by  more  than  1. 

Let  the  roots  of  the  following  numbers  be  found  in  decimals 
carried  to  four  decimal  places. 


144  PURE    EQUATIONS    OF    THE    SECOND    DEGREE       XXXll 


1. 

2. 

7. 

A. 

2. 

27 

8. 

1%. 

3. 

33-75. 

9 

h 

4. 

147307. 

10. 

^6- 

5. 

34f.     '     . 

11. 

l'Q6T' 

6. 

325i|. 

12. 

i^- 

SECTION    XXXII. 

QUESTIONS    PRODUCING    PURE    EQUATION'S    OF    THE    SECOHTS    I>E6R£i£ 

Art.  93.  A  pure  eqitation  of  the  second  degree,  or  a  pure 
quadratic  equation,  is  one  which  contains  the  second  power,  but 
no  other  power,  of  the  unknown  quantity. 

1.  A's  age  is  to  B's  as  7  to  9,  and  the  sum  of  the  second  pow- 
ers of  their  ages  is  1170  years.     Required  their  ages. 

2.  Two  couriers  set  out,  at  the  same  time,  from  two  places 
220  miles  asunder,  and  traveled  towards  each  other  till  they  met ; 
when  it  was  found  that  the  first  had  traveled  only  f  as  fast  as  the 
second,  and  that  the  number  of  hours  they  had  been  on  the  road, 
was  equal  to  the  number  of  miles  the  first  traveled  per  hour. 
Required  the  rate  per  hour  and  the  distance  each  traveled  in  the 
whole. 

3.  A  gentleman  has  two  square  rooms,  the  sides  of  which  are 
as  5  to  6;  and  he  finds  that  it  takes  11  square  yards  more  of 
carpeting  to  cover  the  floor  of  the  larger,  than  it  does  to  cover 
ifiBt  of  the  smaller.  Required  the  length  of  one  side  of  each 
loom. 

4.  A  farmer  had  an  orchard,  in  which  the  number  of  trees  in 
each  row  was  to  the  number  of  rows  as  6  to  5;  and  the  number 
of  bushels  of  apples,  gathered  from  each  tree,  was  to  the  number 
of  rows  as  4  to  5;  moreover,  the  number  of  bushels  in  the 
whole  was  equal  to  80  times  the  number  of  trees  in  one  row 


iiXXII.      PURE    EQUATIONS    OF    THE    SECOND    DEGREE.  145 

How  many  rows  were  there,  how  many  trees  iu  each  row,  and 
how  many  bushels  of  apples  were  gathered  ? 

5  A  gentleman  has  u  rectangular  piece  of  land  50  rods  lona 
and  18  wide,  which  he  wishes  to  exchange  for  another  of  the 
same  area  and  in  a  squure  form.  What  must  be  the  length  of 
one  side  of  the  square? 

6.  A  man  wishes  to  make  a  cistern  containing  800  gallons, 
the  bottom  of  which  shall  be  a  square,  and  the  height  6  feet. 
Required  the  length  of  one  side  of  the  bottom. 

Note.     A  gallon  wine  measure  is  231  cubic  inches. 

7.  An  acre  contains  160  square  rods.  What  is  the  length  of 
one  side  of  a  square  containing  an  acre  of  land  ? 

8.  What  would  be  the  length  of  one  side  of  a  square  contain- 
ing 12  acres? 

9.  What  number  is  that,  to  which  if  10  be  added,  and  from 
which  il  10  be  subtracted,  the  product  of  the  sum  and  difference 
will  be  150? 

10.  The  product  of  two  numbers  is  900,  and  the  quotient  of 
the  greater  divided  by  the  less  is  4,     What  are  those  numbers  1 

11.  There  is  a  house,  whose  breadth  "is  to  its  length  as  5  to 
6,  and  whose  height,  exclusive  of  the  roof,  is  to  its  breadth  as  4 
to  5.  Required  the  dimensions  of  the  house,  supposing  that  it 
takes  2200  square  feet  of  boards  to  cover  the  four  sides. 

12.  A  merchant  bought  two  pieces  of  cloth  of  equal  length ; 
the  one  cost  5  shillings  a  yard  more,  and  the  other  5  shillings  a 
yard  less,  than  the  number  of  yards  in  each  piece.  The  price 
of  the  whole  being  ^136  18s.,  how  many  yards  were  there  in 
each  piece,  and  what  was  the  price  of  each  per  yard? 

13.  A  company  at  a  tavern  found  that  their  whole  bill  was 
$45,  and  that  each  had  to  pay  5  times  as  many  cents  as  there 
were  parsons  in  the  company.  How  many  persons  were  there, 
and  how  much  had  each  to  pay  ? 

14.  There  are  two  numbers,  the  sum  of  whose  second  powers 
is  5274,  and  the  difference  of  these  powers  is  1224.  Required 
the  numbers. 

13 


l46  U  ri:CTED    EQUATIONS    OF    THE  XXXIII 

15.  A  mail  lent  a  certain  sum  of  money  at  6  per  cent,  a  year, 
and  found  that  if  lie  multiplied  the  principal  by  the  number  rep 
resenting  the  interest  for  8  months,  the  product  would  be  $900 
Required  the  principal. 


SECTION    XXXIll. 

AFFECTED    EQUATIONS    OF    THE    SECOND    DEGREE. 

Art.  94:.  The  equations  of  the  second  degree,  ivhich  we 
have  hitherto  considered,  involved  the  second  power  only  of  the 
unknown  quantity.  But,  in  its  most  general  sense,  an  equation 
of  the  second  degree,  with  one  unknown  quantity,  is  composed 
of  three  kinds  of  terms,  viz :  one  kind  containing  the  second 
power  of  the  unknown  quantity  ;  another  containing  the  Jirst 
power  of  the  unknown  quantity ;  and  a  third  composed  wholly  of 
known  quantities. 

Such  are  called  affected  et^uations  of  the  second  degree,  or  af- 
fected quadratic  equations. 

1.  There  is  a  rectangular  field  whose  length  exceeds  its  breadth 
by  8  rods,  and  whose  area  is  180  square  rods.  Required  the 
length  and  breadth. 

Let  X  1=  the  breadth  in  rods ; 
then  X  -|-  8  :=  the  length. 
Hence,.  »2  4_8x=:  180. 
If  we  compare  the  first  member  of  this  equation  with  the  sec- 
ond power  of  z  -|-  a,  which  is  z^  -j-  2  a  x  -|-  a^,  we  see  that  it  con- 
tains two  terms  which  correspond  respectively  to  the  first  two 
terms  of  this  second  power,  viz  : 
x2  =  x2, 
2  a  X  =  8  X.     Hence, 
2a  =  8, 
a=z  4, 


XXXIII.  SECOND    DEGREE.  147 

Now  since  16  corresponds  to  a^,  if  we  add  16  to  bMh  mem- 
bers of  the  equation,  x^  -\-Sxzzz  180,  the  first  member  becomes 
a  perfect  second  power  corresponding  to  x^  ~{-  2  a  x  -\-  a^,  and  we 
have 

x2  +  82;+ 16=  180+16=  196. 

We  now  take  the  root  of  each  member.     The  root  of  the  first 
member  is  x  -|-  4,  for  (z  -(-  4)  (x  -(-  4)  =  a;^  -j-  8  z  -|-  16 ;  and  tha^ 
of  the  second  member  is  14.     We  have  therefore, 

X  +  4=:±14. 
Every  positive  quantity  has  two  second  roots,  one  positive  and 
the  other  negative;  for  the  second  power  of  —  a,  as  well  as  that 
of  -|-  «,  is  -|-  a^.  Therefore,  since  in  an  equation  such  as  x  -j-  4 
=  db  14,  the  value  of  x  is  not  determined  until  the  known  quan- 
tity is  transposed,  and  it  may  happen  that  the  negative  as  well 
as  the  positive  root  will  answer  the  conditions  of  the  question^ 
we  place  the  double  sign  dn  before  the  second  member.  This 
sign  is  read  plus  or  minus. 

In  the  above  equation,  transposing  4,  we  have 
x=.  —  4  dc  14.     Calling  14  plus, 
xz=z  10  rods,  the  breadth  ;  and 
z-|-8=:  18  rods,  the  length.     Calling  14  minus, 
x=:— 18,  andx  +  8=:  — 10. 
The  firsv  value  only  of  x  answers  the  conditions  of  the  que& 
lion.     The  second  value  will  however  satisfy  the  equation ;  for. 
(_  18)2  +  8  (— 18)  =  324  —  144  =  180. 

In  order  to  interpret  the  negative  valae,  we  substitute  —  x  for 
-j-zinthe  original  equation,  and  we  have  x^  —  8xr=180,  .or 
x^x  —  8)=  180.  This  shows  that  x  now  represents  the  longer 
Bi4e  instead  of  the  shorter.  The  solution  of  the  equation, 
jc2  —  8  X  r=  180,  will  be  similar  to  that  of  the  following  question. 
2.  Twenty  times  a  certain  number  exceeds  its  second  powet 
liy  75      What  is  that  number  ? 

Let  X  =  the  number. 
Then,  x^  +  75  =  20  x.     Transposing, 
x2  — 20x.=  — 75. 


i48  AFFECTED    EQUATIONS    OF    THE  XXA.Q1 

In  this  equation  the  term  containing  the  first  power  of  x  being 
negative,  in  order  to  render  the  first  member  a  perfect  second 
power,  we  compare  it  with  the  second  power  of  x  —  a,  which  ia 
x^ — 2ax-\-  a^^  and  we  have 

z2  =  x\  • 

—  2ax  =  — 20  a;.     Hence, 

—  2a    =—20, 

—  a    =  —  10, 
a2  =       100. 

Adding  100  to  each  member,  we  have 

a;2_20x-f  100  =  — 75+100=i25. 
We  now  take  the  root  of  each  member.     The  root  of  the  first 
isz  — 10;  for,  (x  —  10)  (x— 10)  =  2;^— 202  + 100,  and  that 
of  the  second  is  zt  5.     Therefore, 

X  —  1 0  =  ±  5.     Transposing, 
X  =  10  rt  5 ;  hence, 
X  =  15,  or  X  ==  5 
Both  values  of  x  are  positive,  and,  therefore,  both  answer  the 
conditions  of  the  question.     Indeed, 

15.  15  +  75  =  20.  15;  also,  5  .  5  +  75  =  20  .  5. 
Hence  we  see  the  propriety  of  giving  the  double  sign  to  the 
root  of  the  second  member. 

Art.  95.  Any  affected  equation  of  the  second  degree  may  be 
reduced  to  the  form  of  x^-\-px=z  q,  p  and  q  being  any  known 
quantities,  positive  or  negative. 

ll  is  manifest  that  an  equation  may  be  reduced  to  this  form  in 
the  following  manner.  1.  Clear  the  equation  of  fractions  if 
necessary ;  transpose  all  the  terms  containing  x^  and  x  into  tht 
first  member y  and  the  knoion  terms  into  the  second ;  reduce  the 
tttms  which  contain  x^  into  one  term^  and  those  which  contain  % 
into  another ;  also,  reduce  the  known  quantities  in  the  secona 
mejuher  to  one  term.  2.  If  the  term  containing  x^  is  not  positive, 
make  it  so  by  changing  all  the  signs.  3.  If  tlie  coefficient  of  r^ 
vs  not  1 ,  divide  all  the  terms  by  that  coefficient. 


A.XXIII.  SECOND    DEGRL  1.  149 

1.  A  draper  bought  a  quantity  of  cloth  for  c£27      If  he  haj 

bought  3  yards  less  for  the  same  sum,  it  would  have  cost  him  15 

shillings  a  yard  more.     How  many  yards  did  he  buy  ? 

Let  X  =.  the  number  of  yards. 

27 
Then  —  =  the  price  per  yard  in  pounds,  and 

27 

=  the  price  per  yard,  if  he  had  bought  3  yards 


X— 3 

27         27       3 

less.     Hence, r-= \-  -.   Clear  the  equation  of  fractions, 

X  —  o        X       A^ 

108x=  1082  — 324  +  3x2  — 9xj 

tzanspose,  reduce,  and  change  the  signs, 

3  x2  _  9  x  =  324 ;  divide  by  3, 

x2  — 3x  =  108. 

Here  p  =  —  3,  and  q  z=.  108. 

Comparing  the  first  member  with  x^ — 2ax-|-a2,  we  have 

x2  =  x2, 

—  2ax  =  — 3x, 

—  2a    =  —  3, 

—  a    =-t, 

Adfting  f  to  each  member  of  x^  —  3x  =  108,  we  have 
x2  — 3x  +  f  =  l08  +  f  =^1^  +  1  =  ^J 

We  now  take  the  root  of  each  member.  The  root  of  the  first 
is  X  —  1^,  because  (x  —  ^)  (x  —  ^)=ix^  —  3  x  -[-  } :  and  that  of 
the  second  is  db  ^. 

Hence,  x  —  |  =  ±  V"*     Transposing  —  f , 

x  =  f±^=:^=12;  orx  =  — J^  =  — 9. 

The  fir  jt  value  only  of  x  answers  the  conditions  of  the  ques- 
iion 

In  order  to  interpret  the  second  value  of  x,  viz  :  xz=z  — 9,  we 

Bubstitute  — X  for  -{-x  in  the  original  equation,  which  then  be- 

27  27     .   3  27  27    ,    3       . 

comes r= h  T.  or r-^  = +  -j  .    smc(j 

—  x  —  3       _x~4'  x  +  3  X    ^4' 

13* 


/.'>0  AFFECTED    EQUATIONS    OF    THE  XXXIII 

when  the  denominator  is  negative,  it  is  the  same  as  if  the  de- 
nominator were  positive  and  the  whole  fraction  were  preceded 
by  the  sign  — . 

This. last  equation  becomes  by  transposition, 

27  27  3 

—  =z  -\-  - ,  which  answers  to  the  following  question. 

A  draper  bought  a  quantity  of  cloth  for  ^7.  If  he  had  bought 
3  yards  more  for  the  same  sum,  it  would  have  cost  him  15  shil- 
lings per  yard  less. 

Let  the  learner  solve  the  question  as  now  stated. 

2.  Since  p  may  always  represent  the  coefficient  of  the  first 
f ower  of  the  unknown  quantity,  and  q  the  known  term,  let  us 
solve  the  general  equation  x^-\-px=:q,  by  comparing  the  first 
member  with  x^-\-2  ax-\-a^.     We  hare 


X2 

=  xs. 

2az 

=  px, 

2a 

—  p. 

a 

p 

-2' 

flS 

"~4* 

Adding  ^  to  each  member,  we  have 

^^+P^  +  ^'  =  2  +  j- 

We  now  take  the  root  of  each  member.     That  of  the  first 

member  is  a;  +  |,  since  ^z  + 1 W x  + 1 j  =  x^  +p  x  + 1- 

The  root  of  the  second  member  can  only  be  indicated,  until  defi- 
nite valued  are  assigned  top  and  q.     We  have  then. 


+  |  =  ±(<?  +  ^)*;  hence, 


.=-|± 


XXXIII.  SECOND    DEGREE.  151 

Remark.  The  second  root  of  a  quantity  is  properly  expressed 
by  the  exponent  ^.  For  the  second  root  of  a  quantity  multi- 
plied by.  itself,  must  produce  that  quantity.     Now,  according  to 

the  rule  for  exponents,  Art.  19,  o^  .  a^  :=  a-     ^^  =  a^  or  a , 

therefore  the  second  root  of  a  is  a^.    In  like  manner,  the  second 

root  o^  q-\-~-  is  expressed  thus,  zb  (  9  +  ^  )  • 

Art.  96.  From  the  solution  of  the  preceding  general  equa- 
tion, we  deduce  the  following 

RULE  FOK    SOLVING  AFFECTED  EQUATIONS    OF  THE    SECOND    DEGREE. 

1.  Reduce  the  equation  to  the  form  of  x^  -\-px==iq. 

2.  Make  the  first  member  a  'perfect  second  power  y  by  adding  to 
both  members  the  second  power  of  half  the  coefficient  of  x,  (or  of 
the  first  power  of  the  unknown  quantity). 

3.  Extract  the  root  of  each  member.  The  root  of  the  first 
member  will  consist  of  two  terms,  the  first  of  which  is  x,  or  the 
unknown  quantity  y  and  the  second  is  half  the  coefficient  previous- 
ly found,  and  the  root  of  the  second  member  must  have  the  double 
sign  do- 

4.  Transpose  the  known  term  fj  om  the  first  member  to  the  sec- 
ond and  reduce,  and  the  value  of  x  will  be  found. 

Since  p  and  q  may  be  either  positive  or  negative,  it  is  evident 
that  the  general  equation  admits  of  four  forms,  differing  only 
with  regard  to  the  signs  of  these  known  quantities,  viz . 

(1)  x^+px=q)    whence,  i  =  — |±^^  +  ^V. 

(2)  x2—pa;=gr;    whence,  x=:+|±^9  +  ^y. 

(3)  x2-|-pa;  =  — j;  whence,2;  =  — 1±^|-  — yV. 

(4)  x2__^x  =  —  g;  whence,  x  =  +|  rb  f^l qy. 

Art.  97.  These  formulae  for  x  enable  us  to  solve  an  affected 
equation  immediately  according  to  the  following 


152  \FFECTED    EQUAl   ONS    OF     I'HE  XXX  £1 J 


When  the  tquation  is  reduced  to  the  form  ofx^-\-px=iq,  thtk 
unknown  quantity  is  equal  to  half  the  coefficient  of  its  first  power, 
taken  with  the  contrary  sign,  plus  or  minus  the  root  of  the  alge- 
braic  sum,  obtained  by  adding  the  second  power  of  half  that  coeffi 
dent  to  the  known  term. 

1.  Ten  times  a  certain  number  exceeds  its  second  power  by  9. 
What  is  that  number? 

Suppose  X  =  the  number. 

Then,  102=  x2-|-9.     Transpose  and  change  the  signs, 
x2  — 10  2;  =  —  9.     Hence,  by  the  last  rule, 

x  =  5=t  (25  — 9)^=:5ih4=:9;  or  =  1. 

2.  Divide  the  number  20  into  two  parts,  such  that  their  pro 
duct  shall  be  120. 

Let  X  =.  one  part ;  then  will  20  —  x=z  the  other  ;  and 

20x  — x2— 120;  or  x2  — 20x  =  — 120. 
Solving  this  equation,  we  have 

z  1=  10  ±  ( 100  —  120)^  =  10  ±  (—  20)^. 
As  the  second  root  of  a  negative  quantity  is  imaginary,  this 
problem  is  impossible. 

Indeed,  to  generalize  this  question,  let  us  suppose  it  is  re- 
quired to  divide  the  number  p  into  two  parts,  such  that  their 
product  shall  be  q. 

Representing  the  two  parts  by  x  and  p  —  x,  we  have 

px  —  x^=zq;  or  x^ — px=.  —  q.     This  equation  givei 

The  quantity  dz  (^ qj     becomes  imaginary  whenever  q 

pa 
18  grerter  than  ^.     The  greatest  value  that  can  be  given  to  gr, 

without    rendering    the    problem    impossible,    is  ^.       Then 

^ qj    becomes  zero,  and  x,  as  well  as  p  —  x,  is  equal  to  — , 


XXXIII.  SECOND    DEGREE  i53 

The  produit  of  the  two  parts  is  then  eqaal  to  ^.    HencCj  the 

greatest  product  that  can  be  produced  by  multiplying  the  two 
parts  of  a  number  together,  is  the  second  power  of  half  that 
number. 

3.  There  is  a  square  garden,  the  number  of  square  rods  in 
which  exceeds  the  number  of  rods  round  it  by  165.  Required 
the  length  of  one  side 

4.  A  man  built  a  certain  piece  of  wall  for  $27 ;  and  he  found 
that  the  number  of  dollars  he  received  per  rod,  was  6  less  than 
the  number  of  rods  in  the  length  of  the  wall.  Required  the 
number  of  rods  and  the  price  per  rod. 

5.  The  difference  of  two  numbers  is  5,  and  the  sum  of  their 
second  powers  is  15325.     What  are  those  numbers? 

.  6.  A  farmer  bought,  at  $1  per  square  rod,  a  rectangular  field, 
the  length  of  which  was  to  the  breadth  as  5  to  3.  After  having 
built  a  wall  round  it,  which  cost  $2  a  rod,  he  found  that  the  pur- 
chase money,  together  with  the  expense  of  fencing,  amounted  to 
$0640.     Required  the  dimensions  of  the  field. 

7.  A  drover  bought  a  certain  number  of  sheep  for  $50,  and  a 
number  of  calves,  greater  than  that  of  his  sheep  by  3,  for  $52 
Moreover,  the  price  of  a  sheep  and  a  calf  together  was  $9.  Re- 
quired the  number  of  each  kind. 

8.  A  man  haying  traveled  160  miles,  found  that  if  he  had  trav- 
eled one  mile  more  per  hour,  he  would  have  been  8  hours  less 
upon  the  road.  Required  his  rate  of  traveling  and  the  numbe* 
of  hours  he  was  upon  the  road. 

9.  A  jockey  sold  a  horse  for  $150,  and  gained  half  as  mucj 
per  cent,  as  the  horse  cost  him.  How  much  did  the  horse  cost 
him? 

10.  A  man  bought  a  squave  piece  of  land  for  $2  per  square 
foot.  After  having  sold  from  it,  at  the  rate  of  $3^  per  foot,  a 
rectangular  piece,  the  length  of  which  was  equal  to  one  side  of 
ihe  square  and  the  breadth  30  teet,  he  found  that  what  remained 
had  cost  him  only  $4400.  Required  the  length  of  one  side  of 
»he  square. 


154        AFFECTED  EQUATIONS  OF  THE  SECOiND  DEGREE.      XXXIII 

11.  A  grocer  bought  60  lbs.  of  coffee  and  80  lbs.  of  tea  foi 
$46;  but  he  found  that  $1  would  buy  8  lbs.  more  of  cofl^ee  than 
it  would  of  tea.  Required  the  price  of  the  tea  and  coffee  pe. 
pound. 

12.  There  is  a  public  square  whose  ^ide  is  80  rods  long,  sur 
rounded  by  a  walk  of  uniform  breadth,  which  contains  1344 
square  rods.     Required  the  breadth  of  the  walk. 

13.  What  number  is  that  to  which  if  its  second  root  be  added, 
the  sum  will  be  240  ? 

14.  A  father  left  an  estate  of  $30000  to  be  divided  equally 
imong  his  sons ;  but  one  of  these  dying  immediately  after  bis 
father,  the  estate  was  divided  among  those  remaining,  each  of 
whom  leceived  $1500  more  than  he  would  have  received,  if  all 
had  been  living.     How  many  sons  did  the  father  leave  ? 

15.  A  poulterer  had  a  certain  number  of  fowls,  each  of  which 
produced,  during  the  year,  three  times  as  niany  chickens  as  there 
were  fowls ;  and,  at  the  end  of  the  year,  he  found  that  his  whole 
stock,  young  and  old,  was  444.  How  many  fowls  had  he  at 
first? 

16.  Two  men,  A  and  B,  traded  together.  A  put  in  a  certain 
sum  for  4  months,  and  B  put  in  $350  for  2  months.  They 
gained  $99,  and  A  received  for  principal  and  gain  $136.  How 
much  stock  did  A  put  in? 

17.  A  gentleman  has  a  pleasure  garden  80  rods  long  and  60 
rods  wide,  surrounded  by  a  walk  of  uniform  breadth.  The  walk 
contains  576  square  rods.     Required  the  breadth  of  the  walk. 

18.  A  grocer  filled  a  cask  containing  40  gallons  with  wine.  • 
He  then  drew  out  a  certain  quantity  and  filled  up  the  cask  with 
water.  After  this  he  drew  out  the  same  quantity  of  liquid  as  be- 
fore, and  found  that  there  remained  in  the  cask  only  22^  gallons 
of  pure  wine.  Hnw  many  gallons  of  liquid  were  drawn  out 
aach  time? 


KXXIV.  THIRD    ROOTS    OF    NUMBERS.  155 


SECTION    XXXIV. 

EXTRACTION    OF    THE    THIRD    ROOTS    OF    NUMBERS. 

Art.  98.  Find  three  numbers  which  are  to  each  oth«r  as  1, 
2,  and  3,  and  whose  continued  product  is  384. 

Let  X  =  the  first  number  ;  then  2xz=.  the  2d,  and  3ar  •=  t^M 
3d.     Hence,  6x3  =  384. 

In  order  to  solve  this  equation,  .we  divide  by  6,  and  have  x^  = 
64,  or  a;  a:  X  r=  64.  We  see  now  that  x  must  be  a  number,  which, 
multiplied  twice  by  itself,  will  produce  64,  and  we  find  by  trial 
that  X  z=  4. 

The  numbers  required,  therefi^re,  are  4,  8,  and  12. 

The  equation  arising  fi-om  the  preceding  question,  is  called 
an  equation  of  the  third  degree,  because  it  involves  the  third 
power  of  the  unknown  quantity ;  and  the  process  of  finding  the 
first  power  of  a  quantity,  when  the  third  is  given,  is  called  ex* 
trading  the  third  root.  The  third  root  of  any  quantity,  is  that 
quantity  which,  being  multiplied  twice  by  itself,  will  produce  the 
proposed  quantity.  Thus,  4  is  the  third  root  of  64 ;  x  is  the 
third  root  of  x^. 

To  facilitate  the  extraction  of  the  third  roots  of  number^,  we 
shall  give  the  third  powers  of  those  integral  numbers,  which  con- 
sist of  but  one  figure.     They  are  as  follows. 

Roots.     1,2,    3,    4,      5,      6,      7,  .   8,      9.  > 
Third  powers.     1,  8,  27,  64,  125,  216,  343,  512,  729.  5 

The  numbers  in  the  2d  line  are  the  3d  powers  of  those  imme- 
diately over  them,  and  the  numbers  in  the  first  line  are  the  third 
roots  of  those  immediately  beneath  them. 

The  third  power  of  10  =z  1000 ;  that  of  100  —  1000000,  and 
that  of  1000  =  1000000000.  Hence,  the  third  power  of  an  in- 
tegral number  between  10  and  100,  that  is,  of  a  number  consist- 
•ng  of  two  figures,  must  be  between  1000  and  1000000 ;  that  is, 
<t  cannot  contain  less  than  four  nor  more  than  six  figures ,  and 


156  EXTRACTION    OF    THE  XXXIV 

the  third  power  of  an  integral  number  between  100  and  1000,  or 
of  a  number  consisting  of  three  figures,  must  be  between  1000000 
and  1 000000000;  that  is,  it  cannot  contain  less  than  seven  nor 
more  than  nine  figures.  In  like  manner,  it  may  be  shown,  that 
the  third  power  of  a  number  consisting  of  four  figures,  cannot 
contain  less  than  ten  nor  more  than  twelve  figures ;  and  so  on. 

We  can,  therefore,  immediately  determine  how  many  figures 
the  root  of  any  number  will  contain,  by  commencing  at  the  right, 
and  separating  the  number  into  portions,  or  periods,  of  three  fig- 
ares  each.  The  left  hand  period  may  contain  one,  two,  or  three 
figures ;  and  the  root  will  contain  as  many  figures  as  there  are 
periods  in  the  power.  This  separation  may  be  denoted  by  ac- 
cents, as  in  the  extraction  of  the  second  root. 

It  appears  also  from  the  table  given  above,  that,  among  inte- 
gral numbers,  consisting  of  one,  two,  or  three  figures,  there  are 
only  nine  which  are  exact  third  powers ;  consequently,  the  roots 
of  the  intermediate  numbers  cannot  be  obtained  exactly,  although 
they  may  be  approximated,  as  we  shall  see  hereafter,  to  any  de- 
gree of  exactness.  Thus,  the  third  root  of  72  is  between  4  and 
5 ;  the  former  being  nearer  the  true  root  than  the  latter. 

When  a  number  consists  of  no  more  than  three  figures,  pro- 
vided it  is  a  perfect  third  power,  its  root  may  be  found  imme- 
diately by  inspection  or  trial ;  when  there  are  more  than  three 
figured  in  the  power,  its  root  is,  in  some  measure,  obtained  by 
trial,  but  a  rule  may  be  found  which  will  greatly  facilitate  the 
operation. 

Let  us  consider  a  number  of  more  than  three  figures,  as  50653, 

which  is  the  third  power  of  37      Let  a  =  the  tens  and  h  =:  the 

units  of  the  root.     Then,  0+6  =  30  +  7  =  37.     The  thiid 

power  of  a  +  6  is 

a3  +  3a26  +  3a62_j_63. 

By  putting  30  instead  of  o,  and  7  instead  of  6,  we  have 

a3=  (30)3  =  27000, 
3a2  6=:3.  (30)2.7=  18900, 
3a62  — 3.30.(7)2  =  4410. 

63  =(7)3  =  343. 


X.XXIV.         THIRD  ROOTS  OF  NUMBERS.  157 

Hence,  a^-\-Sa^b +  3  ab^ +  b^  =  27000  +  18900  +  4410 
+  343  =  50653. 

Therefore,  the  third  power  of  a  number  consisting  of  tens  and 
units,  contains  the  third  power  of  the  tens,  plus  three  times  the 
second  power  of  the  tens  into  the  units,  plus  three  times  the  tens 
into  the  second  power  of  the  units,  plus  the  third  power  of  the 
units. 

Now  let  it  be  proposed  to  find  the  third  root  of  50653,  that 
root  being  supposed  unknown. 

Operation. 
'       50'653(37.     Root. 

27 

23653(27.     Divisor. 
(37)3=  50653  ~ 
0. 
Separating  the  number  into  periods,  we  see  that  the  root  must 
contain  two  figures,  tens  and  units.     The  third  power  of  the 
tens  can  contain  no  significant  figure  below  thousands;  it  must 
therefore  be  found  in  the  50  (thousands).     The  greatest  third 
power  of  tens  contained  in  50  (thousands),  is  27  (thousands), 
he  root  of  which  is  3  (tens).     Subtract  27  (thousands)  from 
60653,  and  there  remains  23653. 

This  remainder  contains  S  a^  b  -{-  3  a  b^  -\-  b^,  or  three  tinits 
the  second  power  of  the  tens  into  the  units,  three  times  the  tens 
into  the  second  power  of  the  units,  and  the  third  power  of  the 
units. 

If  it  contained  exactly  3  a^  b,  or  three  times  the  second  power 
cf  the  tens  into  the  units,  we  should  find  6,  or  the  units  of  the 
root,  by  dividing  by  3  a^,  or  three  times  the  second  power  of  the 
tens;  but,  as  it  contains  something  more  than  three  times  the 
second  power  of  the  tens  into  the  units,  if  we  divide  by  three 
times  the  second  power  of  the  tens,  our  quotient  may  be  greater 
than  the  proper  unit  figure.  Three  times  the  second  power  of 
the  tens  or  of  30,  is  27  (hundreds),  which  is  contained  in  23653 
14 


158  EXTRACTION    OF    THE  XXXIV 

eight  times.  If  8  be  put  with  the  30  already  found,  it  would 
make  the  root  38.  But  (38)^  —  54872,  which  is  greater  than 
th 3  given  number.  Therefore  8  is  greater  than  the  true  unit 
figure.  We  next  try  7,  and  find  that  (37)3  _  50653.  Hence, 
37  is  the  third  root  of  50653. 

As  a  second  example,  suppose  it  is  required  to  find  the  third 
root  of  13614208. 

Operation. 
43'614'208(352:     Root. 

27 

1st  Dividend  =  166 (27 =  1st  Divisor. 

(35)3  =  42875  .  .  . 
2d  Dividend  =      7392  .  .(3675  .  .  =  2d  Divisor. 
(352)3  =  43614208 

Note.  The  points  in  the  above  operation  are  used  to  show 
the  rank  of  the  figures  which  precede  them. 

Separating  fhe  number  into  periods  of  three  figures  each,  we 
see  that  the  root  must  contain  three  figures,  viz  :  hundreds,  tens, 
and  units.  Let  a  =  the  hundreds,  h  =z  the  tens,  and  c  =  the 
units  of  this  root.  The  third  power  of  a -{-  b  -\- c  =1  a^  -\-Sa^b 
+  3a62_|i53^3a2c-f-6a6c  +  3  62c  +  3ac2  4-36c2  +  c3, 
or  a^  +  3a^b-\-Sab^^b^-{-3(a  +  by2c+S{a  +  b)c^  +  c^; 
for,  3  a2  c  +  6  a  6  c  +  3  62  c  =  3  (flS  -|_  2  a  6  +  62)  c,  or 
3  (a+  6)2c,  and  3  a  c2  +  36  c2  =  3  (a  +  6)  c2. 

We  are  first  to  seek  the  third  power  of  the  hundreds  of  the 
root,  which  must  be  found  in  the  43  (millions).  The  greatest 
third  power  of  hundreds  in  43  (millions)  is  27  (millions),  the 
root  of  which  is  3  (hundreds),  which  we  place  at  the  right.  Sub- 
tracting 27  (millions),  or  a^,  from  the  given  number,  we  have 
for  a  remainder  16614208,  or  3  a2  6  -f  3  a  62  +  63,  &,c. 

Our  next  object  is  to  find  6,  or  the  tens  of  the  root.  The  for- 
cnula  3  «2  5 -|_  3  a  69  _|_  53  gjioyyg^  that,  if  we  divide  by  3  a^^  qj 
thre*»  times  the  second  power  of  the  hundreds,  we  shall  obtain  6. 


XXXIV  THIRD    ROOTS    OF    NUMBERS.  1511 

the  tens'  of  the  root,  or  a  number  a  little  too  great.  But  since 
three  times  the  second  power  of  hundreds  multiplied  by  tens, 
can  produce  no  significant  figure  below  hundreds  of  thousands, 
that  is,  below  the  .sixth  place  from  the  right,  it  is  sufficient  to 
subtract  a^,  or  27  (millions),  from  the  first  period,  and  to  bring 
down,  at  the  right  of  the  remainder,  the  sixth  figure,  that  is,  the 
first  figure  of  the  next  period. 

Taking  then  166,  (16600000),  and  dividing  it  by  three  times 
the  second  power  of  3  (hundreds),  which  is  27,  (270000),  the 
quotient  would  be  6  (tens) ;  but  that  being  found  by  trial  to  be 
too  great,  we  take  5  (tens).  Placing  the  5  (tens)  in  the  root  at 
the  right  of  the  3  (hundreds),  we  raise  35  (tens)  to  the  third 
power,  which  gives  42875  (thousands),  or  a^  -\- 3  a^  b  -\-  S  a  b^ 
+  b^.  This  being  subtracted  from  the  given  number,  43614208, 
leaves  739208,  or  3  (a  +  6)2  c  +  3  (a  +  b)  c^  +  c\ 

Now  in  order  to  obtain  the  units  c  of  the  root,  we  must  evi- 
dently divide  the  remainder  by  S\a  -f-  b)^,  that  is,  by  three  times 
the  second  power  of  35  (tens),  which  is  3675  (hundreds).  But 
since  the  second  power  of  tens  multiplied  by  units,  can  have  no 
significant  figure  below  hundreds,  that  is,  below  the  third  figure 
from  the  right,  it  is  sufficient  to  subtract  the  third  power  of  35 
from  the  first  two  periods,  and  to  the  right  of  the  remainder 
bring  down  the  first  figure  of  the  next  period  to  form  a  dividend. 
This  dividend,  7392  (hundreds),  divided  by  3675  (hundreds), 
gives  2  for  a  quotient,  which  we  place  at  the  right  of  35,  and 
obtain  352  for  the  entire  root.  This  root  raised  to  the  third 
power,  gives  43614208,  showing  that  352  is  the  t-ue  root 
sought. 

In  the  process  just  explained,  it  is  necessary,  after  finding  a 
new  figure  in  the  root,  to  raise  the  whole  root,  so  far  as  it  has 
been  found,  to  the  third  power,  and  subtract  the  result  from  as 
many  of  the  left  hand  periods,  as  there  are  figures  already  found 
in  the  root.  But,  by  considering  the  local  values  of  the  figures, 
we  may  shorten  the  process  of  extracting  the  third  root.  To 
show  the  mode  of  doing  this,  we  shall  resume  the  last  question 


160  EXTRACTION    OF    THE  XXX IV 

Operation.. 


43'614'S 
27 ...  . 
16614. 

508(352. 

..(27.... 

45... 

25  .. 

3175  .  . 

5  . 

=za\ 
=  3a2. 
=  3a6. 

=  62. 

• 

—  b. 

=:3a26  +  3a62_|_63. 

=  3  (a +  6)2. 

=  3(a  +  6)c. 

15875 . 

7392 

08(3675  .  . 

210  . 

4 

369604 

2 

508 

=  3(a+6)2  +  3(a  +  6)c 

+«»■ 

7392 

=  3ra  +  6)2c+3(a  +  6) 

c^  +  A 

0. 

We  proceed,  until  we  find  the  second  figure  of  the  root,  in  the 
manner  already  explained ;  except  that  we  annex  to  the  remain- 
der the  whole  of  the  second  period,  separating  by  an  accent  the 
two  right  hand  figures.  At  this  stage  of  the  process,  we  have 
already  subtracted  the  value  of  a^,  and  our  remainder  with  the 
second  period  annexed,  contains  6  (3  a2  +  3  a  6  +  62)  and  some- 
thing more.  We  next  wish  to  find  the  value  of  6  (3  «2_j_3  ^  6+62), 
and  subtract  it  from  what  remains  of  the  first  two  periods,  after 
the  subtraction  of  a^. 

Now  3a2  =  27,  (270000),  is  our  divisor,  and  3a6=3x 
3  (hundreds)  X  5  (tens)  =  45  (thousands),  is  three  times  the  pro- 
duct of  the  last  figure  found  and  the  preceding  figure  of  the  root  ; 
but  as  6  is  of  the  order  of  units  next  below  a,  the  value  of  3^6 
will  contain  a  significant  figure  one  degree  lower  than  is  found 
in  the  value  of  3  a2  j  therefore  45,  =:  3  a  6,  is  to  be  placed  undci 
27,  =  3a2j  one  figure  further  to  the  right. 

We  now  find  b^  z=:  5  (tens)  X  ^  (tens)  =  25  (hundreds),  and 
as  this  contains  a  significant  figure  one  degree  lower  than  is 


X.XX1V  THIRD    ROOTS    OF    NUMBERS.  161 

found  in  the  value  of  3  a  6,  it  should  be  placed  under  this  last, 
one  place  further  to  the  right. 

These  three  results  being  added  as  the  figures  now  stand,  will 
give  3175  (hundreds),  =  3  a'^ -(- 3  a  6  +  6-,  which  multiplied 
by  5  (tens),  =6,  gives  15875  (thousands),  =z  3a26  +  3a6~»  +  63. 
Subtract  this  product  from  the  dividend,  including  the  two  fig- 
ures separated  from  the  right,  bring  down  the  next  period  to  the 
right  of  the  remainder,  and  we  have  739208,  =^  S  (a -\- b)'^  c -\- 
8(a  +  6)c2  +  c3. 

Separating  the  two  right  hand  figures,  we  t^ke  those  remain- 
ing for  a  dividend,  and  find  3  (a -(-6)^,  =  3  times  the  square  of  35 
(tens),  for  a  divisor;  the  resulting  quotient  is  2,  =  c,  which  we 
place  in  the  root.  We  now  multiply  the  preceding  figures  of  the 
root  by  2,  z=  c,  and  take  three  times  the  product,  which  gives 
210  (tens),  :=  3  (a  +  6)  c,  and  place  the  result  under  the  divisor 
one  figure  further  to  the  right,  under  which,  one  figure  still  fur- 
ther to  the  right,  place  4,  =::  c^;  adding  these  numbers  as  they 
stand,  we  have  369604,  =3{a-{- bf-\- S(a +  b)  c -{-  c^,  which 
•multiplied  by  2,  =  c,  gives  739208,  =3(a-f  6)2c-f  3(a  +  6)c2 
-f-c^.  This  product  subtracted  from  the  last  dividend,  inclu- 
ding the  figures  separated  on  the  right,  leaves  no  remainder ;  the 
work  is  therefore  complete. 

Art.  90.  From  the  preceding  examples  and  explanations  re 
suits  the  following 

RULE    FOR    EXTRACTIJVG    THE    THIRD    ROOTS    OF    NUMBERS. 

1.  Commencing  at  the  right ,  separate  the  number  into  periods 
of  three  ^gures  each  ;  the  left  hand  period  may  contain  one,  two 
iT  three  figures. 

2.  Find  the  greatest  third  power  in  the  left  hand  period, 
place  its  root  at  the  right,  and  subtract  the  potoer  from  thai 
veriod. 

3.  To  the  right  of  the  remainder  bring  down  the  next  period, 
separating  by  an  accent  the  two  right  hand  figures,  and  the  re- 
sult will  form  a  dividend.  For  a  divisor  take  three  times  the 
second  power  of  the  root  already  found.     Divide  the  Mvidend  bn 

14* 


162  EXTRACTION    OF    HIE  XXXIV 

the  divisor,  i  ejecti.  ig  the  two  figures  of  the  former,  hcfoi  e  sep- 
arated, and  put  the  quotient  as  the  second  figure  of  the  root. 

4.  Take  three  times  the  product  of  the  figure  last  found  by  the 
preceding  part  of  the  root,  and  place  it  under  the  divisor,  one 
figure  further  to  the  right ;  under  which,  one  figure  further  to 
the  right,  place  the  second  power  of  the  figure  of  the  root  last 
found.  Add  together  the  divisor  and  the  numbers  placed  under 
it,  as  the  figures  stand,  and  multiply  the  sum  by  the  figure  of  the 
root  last  found.  Subtract  this  product  from  the  dividend,  inclu* 
ding  the  two  figures  separated. 

5.  To  the  right  of  the  remainder,  bring  down  the  next  period, 
forming  a  new  dividend,  in  the  same  manner  as  the  first  was 
formed.  Take  for  a  divisor  three  times  the  second  power  of  the 
whole  root  so  far  as  found;  divide,  rejecting  the  two  right  hand 
figures  of  the  dividend,  and  place  the  quotient  as  the  next  figure 
of  the  root. 

6.  Find  three  times  the  product  of  the  last  figure  by  the  whole 
of  the  preceding  part  of  the  root,  and  place  it  under  the  diviso^ 
one  figure  further  to  the  right;  under  this,  place,  one  figure  fur 
ther  to  the  right,  the  second  power  of  the  last  figure  of  the  root 
found.  Add  the  divisw  and  numbers  placed  under  it,  as  the  fig- 
ures  stand,  TTi^J^'vly  the  sum  by  the  last  figure  of  the  root  found, 
and  subtract  the  pr  oduct  from  the  dividend. 

7.  Repeat  the  operations  stated  in  the  5th  and  6th  parts  of  the 
rule,  until  the  given  number  is  exhausted. 

Remark  I.  If  the  divisor  is  not  contained  in  the  dividend, 
after  the  two  right  hand  figures  have  been  rejected,  put  a  zero  in 
the  root,  and  bring  down  the  next  period ;  the  divisor  for  finding 
the  succeeding  figure  of  the  root,  will  then  be  the  same  as  be^ 
fore,  except  with  the  addition  of  two  zeros  at  the  right. 

Remark  2.  If  the  number  to  be  subtracted  exceeds  that  from 
which  it  is  to  be  taken,  diminish  the  last  figure  found  in  the  root, 
until  a  number  is  obtained  which  can  be  subtracted. 

Art.  lOO.  We  can  always  ascertain  from  the  remainder, 
whether  the  figure  last  placed  in  the  root,  is  so  great  as  it  should 
he.     The  third  power  of  a  is  a^,  and  that  of  a  +  I  is  a^  +  3  a^ 


XXXV  THIRD    ROOTS    OF    NUMBERS  163 

f-3a  + 1.  Hsre  the  roots  differ  by  1,  and  the  powers  differ  by 
3a^+Sa+l.     Hence, 

If  the  remainder  after  subtraction,  contain  three  times  the  sec-^ 
and  power  of  the  root  already  found,  plus  three  times  that  root, 
plus  1,  or  more,  the  last  figure  of  the  root  is  not  sufficiently  great 
by  1  at  least. 

Thus,  m  the  last  example,  if  we  had  taken  4  instead  of  5  for 
the  second  figure  of  the  root,  the  remainder  would  have  bee  a 
4310,  which  exceeds  3  .  (34)2  _f.  3  .  34  ^  i. 

1.  What  is  the  third  root  of  127263527? 
Operation. 
127203'527(503.     Root. 
125 
22^63(75 
22635'27(7500 
450 
9 


754509 

3 

2263527 

0. 

Find  the  third  roots  of  the  following  numbers. 

2.  300763.  6.     37595375. 

3.  59319.  7.    48228544. 

4.  753571.  8.     751089429. 

5.  456533.  9.     27243729729 


SECTION    XXXV. 

THIRD     ROOTS    OF    FRACTIONS  —  AND     THE      EXTRACTION     OF      THIRD 
ROOTS    BY    APPROXIMATION. 

Art.  101.  A  fraction  is  raised  to  the  third  power,  by  multiply- 
ing it  twice  by  itself;  but  as  fractions  are  multiplied  together  by 
lak  ing  the  product  of  their  numerators  for  a  new  numerator,  and 


164  THIRD    ROOTS    OF    FRACTIONS,    AND  XXX \ 

ihat  of  their  denominators  for  a  new  denominator,  it  follows  thai 
a  fraction  is  raised  to  the  third  power  by  raising  both  numeratoi 

and  denominator  to  the  third  power.      For  example,   (y)    = 

a     a    a  a^ 

b  'b  'b   ~~¥ 

Hence,  the  third  root  of  a  fraction  is  founc  by  taking  the  tiiird 
root  of  both  numerator  and  denominator.     Thus,  the  third  root 

1.  Find  the  third  root  of  fff. 

2.  Find  the  third  root  of  ^\^^. 

3.  Find  the  third  root  of  ^-^^j. 

4.  Find  the  third  root  of  if  ^ff 

5.  Find  the  third  root  of  83f  |^f . 

Art.  10^.  When  either  the  numerator  or  denominator  is  not 
an  exact  third  power,  the  root  of  the  fraction  can  be  obtained 
only  by  approximation.  For  example,  if  the  third  root  of  ^  be 
required,  we  may  multiply  both  terms  of  the  fraction  by  49,  the 
second  power  of  the  denominator  7,  and  the  fraction  becomes 
^J.  The  denominator  is  now  a  perfect  third  power,  the  root 
of  which  is  7,  and  the  nearest  root  of  the  numerator  is  5.  The 
approximate  root,  therefore,  of  ^  is  ^  +,  which  differs  from  the 
true  root  by  less  than  j^. 

If  a  more  accurate  root  be  required,  we  may,  after  having  mul- 
tiplied both  terms  of  the  fraction  by  the  second  power  of  the  de- 
nominator, multiply  both  terms  of  the  result  by  any  third  power, 

4  4  .  52 

and  then  find  the  nearest  root.      For  example,  -  =  = — -^  =r 

5  5  .  5-* 

4  .  52  .  123        172800 

-   '  ^^  '  ,^-  =  — — -— ,  the  approximate  third  root  of  which  is 

5.5-.  12"*        5-^  .  12'* 

II  — .     This  root  is  exact  to  within  ^\j,  the  product  of  ^  by  j^ 
What  ns  the  thv^d  root  of  §,  accurate  to  within  i  •  t^  =  tV  ^ 
„,    ^        2       2  .  32       2  .  32  .  153        60750        ^ 
^"  ^"'"  3  =  3732  ""s7PTi5"3^  3^715-3'    '^"    ^°^*    **' 

which  is  }f  -|-. 


K^XXV.        APPROXIMATE    THIRD    ROOTS    OF    NUMBERS.  IG5 

Remark.  To  find  the  third  root  of  a  fraction  to  within  a  given 
limit,  divide  the  denominator  of  the  limit  by  that  of  the  given 
fraction ;  then  multiply  both  terms  of  the  given  fraction  by  the 
second  power  of  its  denominator  and  the  third  power  of  the  quo- 
tient previously  found  ;  after  which  take  the  nearest  root.  Thus, 
in  the  last  question,  ^  is  the  limit.  The  denominator  45  divi- 
ded by  3,  gives  15  for  a  quotient.  We  then  multiply  both  terms 
of  §  by  32  and  153. 

1.  Find  the  third  root  of  f  to  within  ^. 

2.  Find  the  third  root  of  1%  to  within  y^-5. 

3.  Find  the  third  root  of  ^  to  within  y^. 

In  a  similar  manner,  we  may  approximate  the  third  roots  o! 
whole  numbers  which  are  not  perfect  third  powers,  by  convert- 
ing them  into  fractions,  whose  denominators  are  third  powers 

^^     ^      2.123       3456  .    ..  .^.  . 

1  nus  2  =:  =  ,  the  approximate  root  of  which  is  |f  -j- 

exact  to  within  y^. 

Art.  1 03.    But  the  most  convenient  way  to  approximate  the 

third  root  either  of  a  whole  number  or  a  fraction,  is  to  change  it 

into  a  decimal,  whose  denominator  is  the  third  power  of  10,  100, 

or  1000,  &,c.,  and  take  the  root   of  the   result.      Thus,  3=: 

3  .  103       3000    , 

— Tjp—  =  TKrTRi  the  root  of  which  is  |^  +  z=  1*4  -|-.     If  a  more 

accurate  root  is  wanted,  we  may  reduce  3  to  a  fraction  whose 

3     1003 
denominator  is  the  third  power  of  100;  thus,  3=      '  :=. 

\%%%U%.  the  root  of  which  is  |^^  +  =  1-44 -f.  Hence  we 
see  that  three  zeros  are  annexed  for  every  additional  decimal  of 
the  root.  Indeed,  this  is  evident ;  for  the  third  power  of  1  is 
001,  and  the  third  power  of  -01  is  '000001;  thus,  there  are 
three  times  as  many  decimals  in  the  power  as  there  are  in  the 
root. 

We  may  therefore  omit  the  denominator,  and  merely  annex 
three  times  as  many  zeros  to  the  number  as  we  w  sh  to  have 
decimals  in  the  root.     Nor  is  it  necessary  to  add  them  all  a< 


166  PURE    EQUATIONS    OF    THE  XXXV  i 

once,  but  only  to  annex  three  to  the  remainder,  when  a  new  fig* 
ure  of  the  root  is  required. 

In  like  manner,  to  find  the  root  of  a  vulgar  fraction,  convert 
it  into  a  decimal,  with  thrice  as  many  decimals  as  are  required 
in  the  root. 

If  the  number  whose  root  is  sought  contain  integers  and  deci- 
mals, and  the  number  of  decimals  be  not  a  multiple  of  three,  make 
it  so  by  annexing  zeros  to  the  right,  which  does  not  change  the 
value,  but  only  the  denomination ;  or,  point  the  number  botli 
ways  from  the  decimal  point,  and  then  complete  the  right  hand 
period,  if  necessary,  by  annexing  zeros. 

After  these  preparations,  the  third  root  of  a  number  contain- 
ing decimals,  is  found  in  the  same  way  as  that  of  an  integral 
number,  care  being  taken  to  point  off  one  third  as  many  deci- 
mals in  the  root  as  there  are  in  the  power. 

Extract  the  third  roots  of  the  following  numbers,  finding  three 
decimals  in  each  root. 

1.     2. 

Z.     7. 

3.  115. 

4.  15. 

5.  25-7. 

6.  025. 

7.  12-374. 

8.  1256-4. 


SECTION    XXXVl. 

«tUKSTIONS    PRODUCING    PURE    KQUATION8    OF    THE    THIRD    DEORFIC. 

Art.  104.  A  pure  equation  of  the  third  degree  contains  the 
third  power,  but  no  other  power,  of  the  unknown  quantity. 

1.  Three  numbers  are  to  each  other  as  2  3,  and  5;  and  their 
product  is  240.     What  are  these  numbers? 


9. 

h 

10. 

A- 

11. 

7^. 

12. 

12*. 

13. 

3§. 

14. 

A^' 

15. 

m- 

16. 

22^. 

X-XXV''!.  THIRD    DEGREE  167 

2  A  rectangular  box  contains  315  cubic  feet.  The  breadth 
is  I  and  the  depth  is  l  of  the  length.  Required  the  three  dimen- 
sions. 

3.  There  are  two  numbers,  such  that  the  second  power  of  the 
greater  multiplied  by  the  less  is  500,  and  the  second  power  of 
the  less  multiplied  by  the  greater  js  250.  What  are  the  num- 
bers? 

4.  The  depth  of  a  cellar  is  to  its  length  as  4  to  15,  and  the 
breadth  is  to  the  depth  as  1 1  to  4 ;  moreover,  the  cellar  holds 
5280  cubic  feet.     Required  the  three  dimensions. 

5  A  pile  of  bricks  is  8  feet  high,  16  teet  wide,  and  32  feet 
long.  What  would  be  one  of  its  sides,  i^  i»  were  in  a  cubical 
form  ? 

6.  A  gentleman  bought  carpeting,  sufficient  to  cover  the  floof 
of  a  square  room,  for  $54.  The  carpet  cost  per  square  yard 
half  as  many  shillings,  as  there  were  feet  in  one  side  of  the  room 
Required  the  side  of  the  room. 

7.  The  less  of  two  numbers  is  equal  to  one  third  of  the  sum 
of  both ;  and  the  square  of  the  greater  multiplied  by  the  less  is 
964.     What  are  these  numbers  ? 

8.  A  bushel  is  2150f  cubic  inches.  Required  one  side  of  a 
cubical  box,  which  shall  contain  5  bushels. 

9.  The  number  of  cubic  feet  in  a  pyramid  is  found,  by  multi- 
plying together  the  number  of  square  feet  in  the  base,  and  one 
third  of  the  altitude.  If  the  base  of  a  pyramid  is  a  square,  and 
the  altitude  is  four  times  one  side  of  the  base,  what  is  the  alti- 
tude, and  what  is  one  side  of  the  base  of  a  pyramid,  which  con- 
tains 40000  cubic  feet? 

10.  The  solid  contents  of  a  cylinder  are  found,  by  taking  the 
continued  product  of  the  length,  the  square  of  the  radius  of  the 
base,  and  the  number  3-14159.  Required  the  radius  of  the  base, 
and  the  length  of  a  cylinder,  if  the  length  is  to  the  radius  as  7  to 
2,  and  the  cylinder  contains  87-96452  cubic  feet. 

11  The  solidity  of  a  sphere  Is  §  of  314159  multiplied  by  the 
cube  of  the  radius.  Required  the  radius  of  a  sphere,  which  cod- 
tains  28  cubic  feet. 


If»8  POWERS    OF    MONOMIALS.  XXXV f. 


SECTION    XXXVII. 

POWERS    OF    MOJTOMIALS    OR    SIMPLE    ALGKBRAIC    QUANTITIES. 

Art.  lO^.  Any  power  of  a  quantity  may  be  found  by  multi- 
plying  it  by  itself  a  number  of  times  denoted  by  the  index  of  the 
powei  minus  one.  Thus,  the  second  power  of  a  or  a^  is  «  .  «  = 
a  ♦" '  zz:  flfi  X2  =3  a^^  Art.  lO  :  the  third  power  of  a  is  a  .  a  .  a 
.z:;  a*  ■+"  1  +  ^  =:  a^  X  3  -—  q3  .  xhQ  fifth  power  of  a  is  a  .a  .a  .a  .  a 
=:  ^1  +  1  +  ^  +  ^  +  ^  =z  a^  >^^  z=  a^  The  second  power  of  ab  is 
a  6  .  fl  6  z=  fli  X  2  &!  X  2  -,  ^2  ^2  .  the  third  power  of  2  6  c  is 
26c  .26c  .26c  =  21X3  61X3^1X3  —  23  63^3  =^8  63  c3;  and 
the  fourth  power  of  4  6^  c3  d^  is  =  4^  6^x4  ^3x4  ^4x4  _ 
256  6^  c^^  d^^.  In  these  examples,  adding  the  exponent  of  any 
quantity  to  itself,  is  the  same  as  multiplying  this  exponent. 
Hence  we  have  the  following 

RULE    FOR    RAISING    A    MONOMIAL    TO    ANY    POWER. 

Raise  the  numerical  coefficient  to  the  required  poioer^  and  mul- 
tiply the  exponent  of  each  letter  by  the  number  which  marks  the 
degree  of  that  power. 

It  is  moreover  manifest  that  any  power  of  a  product ^  is  the  pro- 
duct  of  that  power  of  each  of  its  factors.  Thus,  the  fourth  power 
of  4  62  c3  d^,  which  is  256  6^  c^^  ^16^  ig  the  product  of  the  fourth 
powers  of  4,  6^,  c3,  and  d^ 

Remark.  With  regard  to  the  signs,  when  the  index  of  the 
power  is  even,  the  power  will  always  have  the  sign  + ;  but  when 
I  he  index  is  odd,  the  power  will  have  the  same  sign  as  the  root 
This  manifestly  follows  from  the  rules  for  multiplication. 

1.  Find  the  2d  power  of  7  a  m^. 

2.  Find  the  2d  power  of  8  6^  c  z*. 

3.  Find  the  2d  power  of  15  a'*  m^  p^, 

4.  Find  the  7th  power  of  2  x'^  y^. 


XXXVIII.  POWERS    OF    POLYNOMIALS.  16^ 

5.  Find  the  13th  power  of  h^  c^  cP. 

6.  Find  the  10th  power  o^^b^c^d^. 

7.  Find  the  mth  power  of  ^j'^g-^x.     Ans.  /»'*'"  gr'  ^x*". 

8.  Find  the  mth  power  of  p"  g-*. 

9.  Find  the  7«th  power  of  2x2  y3.     Ans.  2'"x2'"y3« 

Remark.     In  the  pre.'.eding  example,  since  m  is  indefinite,  ih« 
powei  of  2  must  be  represented  merely. 

10.  Find  the  wth  power  of  3^^  q^. 

11.  Find  the  4th  power  of  —^p^qK 

12.  Find  the  5th  power  of  —  2  x^  y^, 

13.  Find  the  3d  power  of  —  7  a^  62  ^  d. 

14.  Find  the  6th  power  of  —  2  a  m^  n^p^  (p  x  y. 

2  a  4  flSs 

15.  Find  the  2d  power  of  ^.     Ans.  ^-t^ 


16.   Find  the  2d 

.36c 
power  of  ^^^  . 

17.   Find  the  3d 

^5m2n 

18.    Find  the  4th 

P"^^^  "^21^,2x3- 

SECTION     XXXVIII. 


POWERS     OF     POLYNOMIALS. 


Art.  106.  Any  power  of  a  polynomial  may  be  indicated  by 
enclosing  it  in  a  parenthesis,  and  placing  the  index  of  the  power 
over  it  at  the  right.  Thus,  (2  6  4"  ^)^  represents  the  second 
power  of  2  6-|-c.     The  same  thing  may  be  indicated  by  a  vin- 

2 

culum  and  the  exponent,  thus,  2  6  +  c  . 

Powers  thus  indicated  may  be  raised  to  other  powers  in  the 
same  manner  as  simple  quantities,  that  is,  by  multiplying  the 
exponents.       For    example,    the    fourth   power  of  {a-\-  6)3   ig 
15 


170  POWERS    O*     POLYNOMIALS.  XXX  VII J 

(a~^  b)^^'*  z=i  (a-\-by^.  Moreover,  when  several  compounc 
quantities  are  represented  as  multiplied  together,  the  whole  is 
laised  to  any  power  by  raising  each  factor  to  the  power  required. 
Thus,  the  second  power  of  (2  c  +  </)  (3  m  —  nf  is  (2  c  +  df  X 
(Sm  —  w)6  ;  the  third  power  of  2  a  (6  -\-  c)^  (m  -J-  n)^  is 
Sa^(b-\-c)^  (m-\-ny^.  When  some  of  the  factors  are  mono 
mials,  they  should  be  raised  to  the  powei  required,  in  the  man 
ner  already  explained. 

1.  Indicate  the  4th  power  of  6  m — n-\-p. 

2.  Indicate  the   3d   power  of  (b-\-c-{-  d)^. 

3.  Indicate  the  7th  power  of  {^ab-\-ix y)^. 

4.  Indicate  the  13th  power  of  (x  —  y)^. 

5.  Indicate  the   2d   power  of  (a -j- ^)  (« — ^)^- 

6.  Indicate  the  5th  power  of  3(2: — i/Y  [p  —  q)^. 

7.  Indicate  the   3d   power  of  2  (a  -|-  6  +  c)'". 

3.  Indicate  the  4th  power  of  am(c  —  c?)"*  (x-|-y)". 

9.  Indicate  the  mth  power  of  {a-\-b  -{-  c)^. 

10.  Indicate  the  wth  power  of  («+  b)^  (c  —  rf)^. 

11.  Indicate  the  mih  power  of  (2;-f-2y)". 

12.  Indicate  the  mth  power  of  (a  -\-  xy  {n  —  2)*. 


13. 


,       «,  .a-\-b,         /a  +  6V 

Indicate  the  2d   power  of  — ; — ,.     Ans.   I  — ; — ,  ]  . 
c-\-  d  \c-\-d/ 

14.  Indicate  the   3d   power  of  ,_   ,   \.     Ans.   I  ,_  ." .  ) 

b'^-\-n^  Kb^-Yn^J 

6  +  c 

15.  Indicate  the  4th  power  of  — 7—. 

16.  Indicate  the   3d   power  of  ) — — (^. 

{p  +  qf 

17.  Indicate  the  4th  power  of  ^(^^-f^)  (^  — y)_^ 

^  3  (c  -(-  rf)» 

18.  Indicate  the  7th  power  of  .  ,     ,    \^  i    \    ,    .     v, 

^  b  (r -\- sy  (a -\- b -\- cY 

Art.  1 07.  But  if  we  would  have  the  powers  of  polynomials 
in  a  developed  form,  they  may  be  obtained  by  multiplication,  in 
the  manner  of  simple  quantities.      For  example,  (m-\-n)^=: 


XXXVlfl.  POWERS    OF    POLYNOMIALS,  171 

(m  -j-  w)  (m  -f-  n)  ("^  -\- n)  ^^  m^  -\- ^  m^  «  -j-  3  w  7^2  -[-  n^  ,  and 
(26c  +  J)2  =  (26c+<f)(26c+^=:462c2  +  46c<^+c^. 

To  develop  the  expression  a  (6  +  c)^,  we  must  first  find  the 
value  of  (6  +  c)2,  which  is  62-|_26c  +  c-,  and  then  multiply  by 
fl,  which  gives  aV^  -\-^ahc-\-  a<^.  If  the  multiplication  had 
been  performed  before  raising  6  -|-  c  to  the  second  power,  the 
result  would  have  been  a^  52  _|_  2  ^2  5  ^  -[-  «^  c^,  which  is  errone* 
cus. 

1.  Develop  {m  —  x)^. 

2.  Develoi    c{a-\-hf. 

3.  Develop  (a  +  6 -|- c)^  (m  +  »). 

4.  Develop  d^  (x  -(-  y)^. 

5.  Develop  (2c  +  3rf)2. 

6.  Develop  ,— v-\|. 

But  the  finding  of  the  powers  of  polynomials  by  multiplication 
becomes,  when  the  power  is  of  a  high  degree,  exceedingly  te- 
dious ;  and  a  more  concise  and  expeditious  method  has  been  de- 
vised. The  principle  of  this  method  is  called  the  Binomial  The- 
orem, and  was  discovered  by  Sir  Isaac  Newton.  It  is  primari  y 
applied  to  binomial  quantities,  but  may  be  extended  to  polyno- 
mials. 

Art.  108.  As  a  table  of  the  powers  of  a  binomial  will  some- 
times be  found  convenient  for  the  purpose  of  reference,  we  sub- 
join a  few  of  the  powers  of  a  -|-  t,  obtained  by  multiplication. 

(a-\-xy  ■=.a  -\-x. 

( a  +  Z  )3  =  a3  _|_  3  «2  2;  _|.  3  ^  2;2  _|_  2-3 . 

( a  -f  2^  )^  =  "'^  +  4  a^  a;  -j-  6  a2  a;2  -|-  4  a  x3  +  24 . 

(a-f  x)5  — aS-f  5a4a;4-I0a3x9+10a2  3:3-(_5az4_|_2;5. 

If  the  second  term  of  the  binomial  is  negative,  the  powers  will 
be  the  same  as  when  it  is  positive,  except  that  the  successive 
terms  will  be  alternately  positive  and  negative ;  that  i&,  all  the 
terms  in  which  an  odd  power  of  the  negative  term  enters,  will  be 
negitive,  all  the  others  being  positive.     This  follows  from  the 


172  BINOMIAL    THEOREM.  XX XI A 

rules  for  multiplication ;  because,  when  the  number  of  negative 
factors  is  even,  the  product  is  positive,  but  when  the  number  of 
negative  factors  is  odd,  the  product  is  negative.  The  first  five 
powers  of  a  —  x,  therefore,  are  as  follows,  viz  : 

(a  —  xY  =  a  — X. 

( a  —  a: )  3  2=  a3  _  3  ^2  x  4- 3  a  z2  —  x3^ 

(a  —  iYz^a'^  —  4:a?x-\-Qa^x^  —  4.ax^-\-x\ 


SECTION    XXXIX. 


BINOMIAL     THEOREM. 


Art.  109.  The  most  concise  demonstration  of  this  theorem  is 
that  of  indeterminate  coefficients ;  and,  as  subsidiary  to  the  demon- 
stration, we  shall  prove  the  following  proposition,  viz : 

Ify  whatever  be  the  value  of  x,  (any  indefinite  quantity )y  two 
polynomials  involving  successive  powers  of  x,  as  A  -j-  B  a;  -[-  C  z^ 
+  D  z3  -|.  E  x4,  4.C.,  and  A'  -\- B' x  +  C  x^  +  D'  x^  +  E'  x\ 
Sfc.y  are  equal,  we  shall  always  have  A  =  A',  B  r=  B',  C  =  C, 
D  =  D',  E  =  E',  S^c. ;  that  is,  the  terms  which  do  not  contain  x 
are  equal,  as  are  also  the  coefficients  of  the  same  powers  of  x. 

Since,  in  the  equation  A-\-Bx-\-  Cx^  -f-  D  x^,  &c.,  =  ^'  -|* 
B' X -\- C  x*^ -\- D' x^ ,  &c.,  the  two  members  are  equal,  inde- 
pendently of  X,  they  must  be  equal  when  m=0;  but,  in  this 
case,  the  terms  all  vanish  except  the  first  in  each  member,  and 
Ihe  equation  becomes  A  =  A'.  Subtracting  these  equal  quan- 
tities, and  dividing  the  remainders  by  x,  we  have 

B-^Cx-\-Dx'^,&LC.=B'-^C'x-{-  D'  x9,  &c. 

Again  suppose  z  r=:  0,  and  this  last  equation  becomes  B  z=  B 
fn  like  manner,  it  may  be  proved  that  C=  C\  IP=z  D',  &-c. 


XXXTX.  niNOMIAL    THEOREM.  .173 

Art.  no.  The  binomial  x-\-a  may  be  put  under  the  foim 
of  x^l -f^Y  so  that  (x-f- «)"*==  x'»^l  +  ^y,  Art.  905.     To 

avoid  fractions,  put  y  =  -,  and  by  substitution  we  have  (x-j-a)" 

X 

=  x"*  (1  -j-  y)"* ;  hence,  to  obtain  the  value  of  the  mth  power  of 
x-j-a,  it  will  be  sufficient  to  find  that  of  (l-j-y)*"  developed, 
then  restore  the  value  of  y,  and  multiply  the  whole  by  x**. 

From  the  manner  in  which  a  binomial  is  raised  to  any  power 
by  actual  multiplication,  it  is  manifest  that  (I  +y)'"  developed, 
will  be  of  the  form  o^A  +  By-^  Cy^  +  D  y^ -\- E y"^ ,  &c.,  in 
which  the  values  of  A  and  of  the  coefficients  J5,  C,  D,  &,c.,  as 
well  as  the  number  of  the  terms,  are  wholly  independent  of  the 
value  of  y,  and  are  determined  entirely  by  that  of  the  exponent 
m.  To  make  this  more  evident,  we  subjoin  a  few  of  the  powers 
ofl  +  t,. 

(l-hy)i  =  l+y. 

(l+y)3=l+3y.+  3y2  +  3/3. 

(l+y)5=:l+5y  +  10y2_|_  103,3  _|.5y4  +  3^5. 

We  see,  therefore,  that  in  each  power  A  =.  1,  whereas  J5,  the 
coefficient  of  the  second  term,  is  different  in  different  powers ; 
the  same  is  the  case  with  C,  &c.,  exc«*pt  with  regard  to  the  co- 
efficient of  the  last  term,  which  is  always  1.  Moreover,  in  each 
power  the  number  of  terms  exceeds  by  1  the  number  which 
marks  the  degree  of  that  power.  Hence  we  infer  that  in  the  7»th 
power  there  will  be  m  -|-  1  terms. 

Art.  111.    Suppose,  then, 

[1]     {\+yY  =  A+By  +  Cy^  +  Dy^  +  Ey^,&.<,., 
in  which  the  values  of  yl,  J5,  C,  &,c.  are  to  be  determined. 

We  have  already  inferred  that  A  is  always  1 ;    this  however 
may  be  demonstrated ;  for,  since  equation  [1]  is  true  for  all  val- 
ues of  y,  it  is  true  when  y  ==  0,  which  reduces  the  equation  tc 
15* 


174  BINOMIAL    THEOREM.  XXXIX 

(1)"*=^.  But,  since  every  power  of  1  is  1  we  necessarily 
have  I  ±z  A. 

We  proceed  now  to  investigate  the  other  coefficients  B,  C,  D, 
fcc.  Since  these  coefficients  are  entirely  independent  of  the 
value  of  y,  we  have,  in  like  manner, 

[2]     {\+zY  =  A-[-Bz+Cz^  +  Dz^  +  Ez\&.c. 

Hence,  by  subtraction  and  the  union  of  terms  which  have  a 
common  coefficient, 

[3]     {l+yT  -(l+zT  =  B  [y-z)  +  C  {y^-z^)  + 

D  (y3  __  23)  _|.  £  (3^4  _  ^)^  &,c. 

Each  of  the  factors  y  —  z,  y^  —  ^^^  &lc.j  is  divisible  by  y — z\ 
actually  dividing,  therefore,  the  second  member  of  equation  [3] 
by  y  —  z,  and  representing  the  division  of  the  first  member,  we 
have 

D{y^  +  yz  +  z^)  +  E{y^-\-y^z-\-yz^  +  z^l&LC. 
Add  1  —  1  to  y  —  2,  which  does  not  change  its  value,  and  it 
becomes  y-\-l  —  z  —  I,  or(l  +y)  —  (1  +  «).    The  first  mem- 

ber  of  equation  [41  then  becomes  \^.     . )^    /     /  .     The 

division  can  now  be  performed,  and  gives.  Art.  40, 

...  +  (1 +«)-'. 

Suppose  y  •=.Zy  and  substitute  y  for  z  in  the  second  member 
of  equation  [5]  ;  the  terms  then  become  alike,  and,  as  the  num- 
ber of  them  is  m,  the  sum  of  the  whole  is  tw  (I  -(-  y)*""^.  Sub- 
stitute this  instead  of  the  first  member  of  equation  [4],  and,  in 
the  second  member,  put  y  instead  of  z  and  reduce,  and  we  have 

[6]     m  (1  +  y)-"-!  =  J?  4-  2  Cy  +  3  Dy^  +  ^Ey^,  &c. 

Mu.tiply  both  members  by  1  +y,  arrange  the  second  member 
of  the  result  according  to  the  powers  of  y,  and  equation  [6]  be 
9omes 


XXXIX.  BINOMIAL    THEOREM.  175 

[7]  m(l+yr  =  B  +  (B  +  2C)y  +  {2C+dD)2/^  + 
(3 1>  +  4  ^)  3/3,  &,c. 

By  substituting  now  for  (1  4-^)*",  its  value,  given  in  equatioa 
[1],  and  putting  1  instead  of  Aj  equation  [7]  becomes 
m{l+Bi/  +  Cy^  +  Di/^  +  Ei/,6LC.)  =  B  +  (B  +  2C)y  + 
(2  C+  SD)y^  +  (3D  +  iE) yS,  &c.,  or, 

[8]  m-{-mBi/  +  mC2/^  +  mDi/^+mEi/'^,&,c.  =B  + 
(B  +  2C)y  +  (2C+3D)3/2  4.(3  2>4.4£)y3,&c. 

But,  as  was  proved  at  the  commencement  of  this  section,  the 
terms  not  involving  y  are  equal,  as  are  also  the  coefficients  of  the 
same  powers  of  y. 

Hence,  B=im; 

B  +  2C=mB;  hence,  C=^^)  =  ^^^li^) ; 
iC+3Z>..mC;  hence, Z>=g(^).^^-y(^^^^ 

'iD  +  iE  =  mD;  hence,  E=  :5i^=l)  -- 

m  (m—  1)  (m— 2)  (ot  — 3) 

1.2.3.4 

These  results  are  sufficient  to  enable  us  to  continue  the  forma 

ti  m  of  the  coefficients  as  far  as  we  please.     The  next  coefficient 

ij      -J      1    1     ntlm  —  l)(m  —  2)  (m — S)(m  —  4)         ,    , 
would  evidently  be  — ^ —. — _    J  -r— e~^- -,  and  the 

succeeding  one  "^C^.-!)  (m-2)  (^-3M«»-4)  (^-5)^ 

Substitute  these  values  of  ^,  B,  C,  «fcc.  in  equation  1],  and  A 
becomes, 


{m-l){m-^){m-S) 

1.2.3.4  ^  '  *-°- 


Restoiing  the  "alue  of  y,  viz  :  y  =  -,  we  have, 


176  BINOMIAL    THEOREM.  XXXIA 


(■  '■?•)"= 


.    ,    m       a       m(m  —  1)       a^ 


m(m  —  l){m  —  2)     q3       m(m  —  \)(m  —  2)(m  —  S)     a^ 
1.2.3  *  x3~r  1.2.3.4  '  1^' 

Multiplying  both  members  by  z'", 

^i(m— l)(m  — 2)      x^a^       m(m--l)(m  — 2)(m  — 3) 
'^  1.2.3  a;3^"  1.2.3.4  ^ 

Reducing  the  fractions  to  lower  terms, 

(x  +  a)- =  2»  +  m  X— 1  «  4  ^?i^-=ll)  z^-s  a2  _^ 

1  .  Z 

m(m— ])(m  — 2)  ,       ^^  (;;,— 1)  (;;^_2)  (m  —  3) 

1.2.3  "^  1.2.3.4  ^ 

x"'-'*a\&,c. 

Art.  lis.  Such  is  the  formula  for  any  power  of  a  binomial ; 
from  which  we  readily  deduce  the  law  both  of  the  letters  and  the 
coefficients. 

First,  with  regard  to  the  letters,  we  see  that,  in  the  first  term, 
J,  which  is  the  first  term  of  the  binomial,  is  raised  to  the  power 
to  which  the  binomial  was  to  be  raised,  and  that  the  powers  of 
7  in  the  successive  terms  go  on  decreasing  by  unity. 

Secondlv>_ry.  the  second  term  of  the  binomial,  is  found  in  the 
second  term  of  the  power  with  1  for  its  exponent,  and,  in  the 
successive  terms,  the  powers  of  a  go  on  increasing  by  unity. 

Moreover,  the  sum  of  the  exponents  of  x  and  a  in  the  same 
term,  is  always  equal  to  m,  the  exponent  of  the  power  to  which 
the  binomial  is  raised. 

With  regard  to  the  coefficients ;  we  perceive,  that  the  coeffi- 
cient of  the  first  term  is  1 ;  that  of  the  second  term  is  equal  ia 
•»»,  the  exponent  of  the  power  to  which  the  binomial  is  raised. 


XXXIX.  BINOMIAL    IHEOREM.  L71 

To  obtain  the  coefficient  of  the  third  term,  we  multiply  t^at:' 

m  —  1 
of  the  second,  which  is  m   by  — - — ;  that  is,  we  multiply   by 

m  —  1  and  divide  by  2. 

To  obtain  the  coefficient  of  the  fourth  term,  we  multiply  mat 

m 2 

of  the  third  by  — ^— ;  that  is,  multiply  by  wi  —  2  and  divide  by 
o 

3,  and  so  on 

Art.  113.    Hence,  having  one  term  of  any  power  of  a  bino- 

m  al,  the  succeeding  term  may  be  found  by  the  following 

RULE. 

Multiply  the  given  term  by  the  exponent  of  x.  in. that  term ^  that 
is,  by  the  exponent  of  the  first  or  leading  quantity  of  the  bino- 
mial,  and  divide  the  product  by  the  number  which  marks  the  place 
of  the  given  term  from  the  first  inclusive  ;  diminish  the  exponent 
of  X  by  1,  and  increase  that  of  a  by  1. 

The  coefficient  of  the  first  term  always  being  1,  and  that  of 
the  second  being  the  same  as  the  index  of  the  power  required, 
we  can,  by  the  preceding  rule,  write  any  power  of  a  binomial. 

Let   it   be   required,  for  example,  to  find  the  9th  power  of 

The  first  term  is  x^ ;  the  second  is  9x^  a  ;  the  third  is  found 
Dy  multiplying  9,  the  coefficient  of  the  second,  by  8,  the  expo- 
nent of  x  in  the  same,  dividing  the  product  by  2,  which  marks 
the  place  of  the  second  term,  diminishing  the  exponent  of  x  and 
increasing  that  of  a  each  by  unity.  The  third  term  then  is 
9.8 


2 

36.7 


c?  a2  =3  9  .  4  z7  a2  =  36  x'^  n^.      The    fourth   terra    is 


x^  a^  =  12  .7  x^  a^  z=.  84  x^  a^.      Finding,  in  a  similar 


manner,  the  succeeding  terms,  we  have 

(x  +  a)9  =:i-9  +  9  x8  «  -f  36  a:7  a^  +  84  x^  a^  +  126  x^  a^  -f 

126  x4  aP  +  84  2:3  ^6  _|_  36  ^2  ^7  _|_  9  j;  ^8  _|_  ^9, 

Since  any  quantity  with  zero  for  an  exponent  is  1,  we  may 
vuppose  a^  to  enter  into  the  first  term,  and  x^  into  the  last      If 


178  BINOMIAL    THEOREM.  XXXIX 

we  should  attempt  to  find  another  term  succeeding  a^  or  x^  a®, 
we  should  obtain  for  its  coefficient  — ^  =  —  =  0.  No  addi- 
tional terms  therefore  can  be  obtained. 

Applying  the  rule  and  remembering  that  odd  powers  of  nega- 
tive quantities  are  negative,  we  have  also 

(a  — .  6)10  =  «io  _  10  «9  6  _j_  45  ^8  ^,2  _  120  «7  ^3  _|.  210  a6  54  _ 
252  a^s  +  210  «4  £6  _  120  a3  ^7  _^  45  ^2  68  _  10  a  69 -j_  ftio. 

From  the  preceding  examples,  as  well  as  from  the  table  of 
'powers  given  in  the  Art.  108,  we  infer, 

1.  That  the  number  of  terms  in  each  power  of  a  binomial  ex- 
ceeds by  1  the  index  of  that  power.  Thus,  in  the  fifth  power, 
there  are  six  terms ;  in  the  ninth  power,  there  are  ten  terms. 

2.  When  the  number  of  terms  is  odd,  there  is. one  coefficient, 
in  the  middle  of  the  series,  greater  than  any  of  the  others ;  but, 
when  the  number  of  terms  is  even,  there  are  two  coefficients  in 
the  middle,  of  equal  value  and  greater  than  any  of  the  others. 
Moreover,  those  which  precede  and  those  which  succeed  the 
greatest  or  greatest  two,  are  the  same,  only  arranged  in  an  in- 
verse order. 

Therefore,  when  half,  or  one  more  than  half  of  the  coefficients 
have  been  found,  the  others  may  be  written  down  without  the 
trouble  of  calculation. 

1.  Find  the  seventh  power  of  a -|- 6. 

2.  Find  the  tenth  power  of  x  -\-  y. 

3.  Find  the  fifth  power  of  m  —  n. 

4.  Find  the  eleventh  power  of  6  -|-  c. 

5.  Find  the  thirteenth  power  of  2;  —  y 

6.  Find  the  sixth  power  of  3  a  -|-  6. 

In  this  last  example,  the  numerical  coefficient  of  a  must  be 
raised  to  the  requisite  powers  by  multiplication. 

First  write  the  power,  merely  indicating  the  operations  with 
egard  to  3  a,  and  we  have 

(3«)6  +  6  (3  af  6  +  15  {^af  62_|_29  (3  afb'^+l^  (3a;,2  6< 
+  6  (3  a)  65  -j-  66. 


X-XXIX.  BINOMIAL    THEOREM.  170 

Raising  3  a  to  the  several  powers  indicated,  and  substituting 
the  results, 

729  a6  _|_  6  .  243  a5  6  _(_  15  .  81  a*  b^  +  20  .  27  a^  h^  + 
15  .  9  a2  64  _|.  6  .  3  a  65  _^  66. 

Performing  the  multiplication,  we  have  for  the  final  result, 

729  a6  +  1458  a5  6  +  1215  o^  62  +  540  a?  P  +  135  a^  b*  + 
8a65_|_66. 

7.  Find  the  fifth  power  of  x  +  2  y.  • 

8.  Find  the  third  power  of  6  a  +  5  x.       ' 

9.  Find  the  fourth  power  of  a-{-b  —  2  c. 

When  a  quantity  containing  several  terms,  as  a-j-b  —  ^  ^  Iz 
to  be  raised  to  a  power,  it  is  convenient  to  sfLj!titrte  cthe:-  '.ot- 
ters, so  as  to  render  the  quantity  a  binbmial,  *aioO  this  binonlaJ. 
to  the  required  power,  and  then  restore  the  value  of  the  Ict*'^f& 
substituted. 

Thus,  in  the  present  example,  let  6  — 2  c  :=  m  ;  then  a  -|-  6  - 
2  c  =  a  +  wi.  Now  (a  +  m)*  ==  a^  +  4  a^  m  +  6  a^  rr?  -j- 
4  a  w^  4"  ^^'     ^^^» 

mz=:b  —  2  c; 

m2  =  ( 6  —  2  c)2  =  62  —  4  6  c  +  4  c2 ; 

m3  z=  (6  —  2  c)3  =  63  —  3  62  (2  c)  +  3  6  (2  c)2  —  (2  c)3,  or, 

;;j3  —  ^3  _  6  62  c  +  12  6  c2  —  8  c3  ; 

m4  ==  (6  —  2  c)4  =  64  —  4  63  (2  c)  -f  6  62  (2  c)2  —  4  6  (2  c)^ 
+  (2c)4,or, 

i»4  —  64  —  8  63  c  +24  62c2  — 32  6c3+  ICc^. 

Putting  these  values  instead  of  m,  ff|2,  &c.,  and  performing 
the  miiltiplication  by  4  a^,  6  a^,  &c.,  we  have 

(0  +  6  — 2c)4z=za4_j.4a3  6  — 8a3^.  +  6«2  62__24a3  6c-f 
24a2c2+4a63— 24a62c+48a6c2_32ac3+64— 863c 
+  24  62  c2  —  32  6  c3  +  16  c^. 

10.  Find  the  fifth  power  of  2  a  +  3  x. 

11.  Find  the  third  power  of  4  x  —  3  y  -)-  a. 

12.  Find  the  third  power  ofa-j-64-c-|-d 


180  NUMERICAL    ROOTS    TO    AN  IT    DEGREE.  XL 

Let  a  -\-  b  z=  m,  and  c  -\-  d  =:  n;  then  a  -f-  ^  -f  c  -f-  «?  =s 
m  -\-n. 

13.    Find  the  sixth  power  of  a  -|-  2  6  —  c. 

1  i.    Find  the  fifth  power  ofa  +  6  —  2c  —  3<f. 

In  this  example,  let  «  +  6  rr  m,  and  2  c  -f-  3  rf=rn  j  then  a  i- 

i /J  C 3  £?  I=z  ?yi 71. 


SECTION    XL. 

nOOTS    OF    NUMBERS    TO    ANT    DEGREE, 

Art  114  The  different  powers  of  a  binomial  suggest  thu 
means  of  extracting  roots  to  any  degree,  both  of  numerical  and 
literal  quantities. 

Let  it  be  required,  for  instance,  to  find  the  fifth  root  of 
9765625 

Operation, 
97'65625         (25  =  a  +  6 

32 =  a\ 

656  (80=5a4. 

9765625  —  (25)5  z={a  +  6)5. 
As  the  fifth  power  of  10  is  lOOOOO,  consisting  of  six  figures, 
and  that  of  100  is  10000000000,  consisting  of  eleven  figures,  the 
fifth  root  of  9765625  must  be  between  10  and  100;  that  is,  it 
must  consist  of  two  figures,  tens  and  units.  Let  a  represent  the 
tens  and  b  the  units  of  the  root.  The  formula  for  the  fifth  power 
of  a  binomial  is  {a  + b)^  =  a^  +  ^a"^  b-\-  10 a^ b^,  &c.  This 
shows  us  that  we  are  first  to  seek  the  fifth  power  of  the  tens, 
which  must  be  found  in  the  97,  (9700000) ;  or,  what  is  the  same 
thing,  we  are  simply  to  seek  the  greatest  fifth  power  in  97.  Now 
25  =  32,  and  35  =z  243.  The  greatest  fifth  power,  therefore,  in 
97  is  32,  the  root  of  which  is  2.  Place  2  as  the  first  figure  or 
lens  of  the  root,  subtract  32  from  97,  to  the  remainder  annex  tbfl 
rest  of  the  figures,  and  we  have  6565625. 


XL.  NUMERICAL  ROOTS  TO  ANY  DEGREE.  l&l 

Th.s  remainder  contains  5  a'^  b -\-  10  a^  ^2^  &,c.,  or  five  times 
the  fourth  power  of  the  tens  into  the  units,  and  something  more. 
If,  therefore,  we  divide  by  five  times  the  fourth  power  of  the 
tens,  the  quotient  will  be  the  units,  or  a  number  a  little  too 
great.  But,  as  the  fourth  power  of  tens  into  units,  can  contain 
no  significant  figure  below  the  fifth  from  the  right,  it  is  sufficient 
after  having  subtracted  32  from  97,  to  bring  down  6,  the  next 
figure,  to  the  right  of  the  remainder,  and  to  take  656  for  oui 
dividend.  Five  times  2^  =  80,  which  is  contained  in  656  eighi 
times.  But  if  we  put  8  in  the  root,  at  the  right  of  the  2,  and  raise 
28  to  the  fifth  power,  the  result  will  be  greater  than  9765625. 
The  same  would  be  the  case  with  27  and  26.  But  if  we  take  5 
as  the  unit  figure,  we  find  (25)^  =  9765625.  Therefore  25  is 
the  true  root. 

If  there  were  more  than  ten  figures  in  the  given  number,  there 
would  be  more  than  two  in  its  root.  We  should,  in  that  case,  let 
a  at  first  represent  the  highest  order  of  units  and  b  all  the  rest, 
until  we  found  the  second  figure  of  the  root ;  after  which  a  might 
represent  the  two  figures  found  and  6  the  rest,  and  so  on. 

Moreover,  it  is  easy  to  see,  that  the  number  is  to  be  separated 
into  periods  of  five  figures  each,  except  that  the  left  hand  period 
may  contain  less  than  five;  that  the  root  will  contain  as  many 
figures  as  there  are  periods ;  that  the  fifth  power  of  the  first  fig- 
ure is  to  be  subtracted  from  the  first  period,  the  fifth  power  of  the 
first  two,  from  the  first  two  periods,  that  of  the  first  three,  from 
the  first  three  periods,  &,c. ;  and,  that,  in  each  case,  we  are  to 
take  the  remainder  with  the  first  figure  of  the  next  period  for  a 
dividend,  and  five  times  the  fourth  power  of  the  figures  already 
found  for  a  divisor. 

Similar  explanations  might  be  given  for  the  extraction  of 
fourth,  sixth,  seventh  and  other  roots.  The  mode  of  procedure, 
m  each  case,  may  readily  be  deduced  from  the  formulae. 

Art.  115.    We  may,  therefore,  take  the  general  formula  for 
the  binomial  theorem,  and  deduce  from  it  a  rule  for  extracting 
toots  of  any  degree  whatever.     This  formula  is 
16 


182  NUMERICAL  ROOTS  TO  ANY  DEGREE.  XL 

«*"  -|-  wi  a'"~i  6  -| — ^r — -  a*"~  2  J2^  ^Q^  in  which  m  denoteg 

the  degree  of  the  root  to  be  found.  The  first  two  terms  alone 
in  connection  with  inferences  which  are  easily  drawn  from  what 
precedes,  determine  the  rule,  which  is  as  follows. 

RULE    FOR    EXTRACTING    THE    7»TH    ROOT    OF    A    IfUMBER 

1.  Beginning  at  the  right,  separate  the  number  into  periods 
of  m  figures  each;  the  left  hand  period  may  contain  from  one  to 
m  figures. 

2.  Find  the  greatest  mth  power  in  the  left  hand  period,  and 
put  its  root  at  the  right  of  the  given  number,  as  the  first  figure 
of  the  required  root.  Subtract  the  mth  power  of  this  figure  from 
the  first  period,  and  to  the  right  of  the  remainder  bring  down  the 
first  figure  of  the  next  period  to  form  a  dividend. 

3.  For  a  divisor,  take  m  times  the  (m  —  \)th  power  of  the  root 
already  found.  Divide  and  place  the  quotient  as  the  second  fig' 
ure  of  the  root. 

4.  Raise  these  two  figures  to  the  mth  power,  and  if  the  result 
does  not  exceed  the  first  two  periods  of  the  number,  subtract  it 
from  these  two  periods,  and  to  the  remainder  annex  the  first  fig' 
ure  of  the  succeeding  period  to  form  a  new  dividend.  But,  if  the 
mth  power  of  the  first  two  figures  exceeds  the  corresponding  peri- 
ods,  diminish  the  second  figure  of  the  root,  until  an  mth  power  is 
obtained  which  can  be  subtracted. 

5.  For  a  new  divisor,  take  m  times  the  (m  —  \)th  power  of  the 
whole  root  already  found.  The  division  will  enable  us  to  find  the 
third  figure  of  the  root.  Then  raise  the  three  figures  to  the  mth 
power,  and  subtract  the  result  from  the  first  three  periods  ;  and 
thus  proceed  until  all  the  periods  have  been  used. 

Remark  \st.  It  is  manifest  that  the  second  and  third  roots 
may  be  extracted  according  to  the  above  rule,  as  well  as  accord- 
ing to  the  rules  previously  given.  The  particular  rules  are 
preferable,  only  because  they  render  the  operations  shorter  than 
the  general  rule  would. 

Remark  2rf.      When  the  number  expressing  the  degree  of  the 
rt    can  be  separated  into  factors,  this  may  be  done,  and  wc5 


XLI  ROOTS    OF    MONOMIALS  183 

may  find  success. vely  roots,  the  degrees  of  which  are  denoted  b^ 
these  factors.  Thas,  instead  of  finding  the  fourth  root  imme- 
diately, we  rfiay  first  find  the  second  root,  and  then  the  second 
root  of  that  result.  For  example,  the  second  root  of  a^  is  a^, 
and  the  second  root  of  a^  is  a.  In  like  manner,  to  obtain  the 
sixth  root,  first  find  the  second  root,  and  then  the  third  root  of 
that  result.  To  obtain  the  eighth  root,  extract  the  second  oot 
three  times,  and  to  get  the  ninth  root,  extract  the  third  root 
twice. 

1.  Find  the  fourth  root  of  625. 

2.  Find  the  fourth  root  of  20736. 

3.  Find  the  fourth  root  of  28398241. 

4.  Find  the  fifth  root  of  2073071593. 

5.  Find  the  fifth  root  of  41-8227202051. 

Remark.     Point  off  both  ways  from  the  decimal  point. 

6.  Find  the  sixth  root  of  4826809. 


SECTION     XLI. 

ROOTS    OF    MOfTOAllALS    OR    SIMPLE    ALOCBRAIC    QUAXTTTISS. 

Art.  116.  From  the  method  given  in  Art.  105,  for  obtain- 
ing powers  of  monomials,  results  the  following 

RULE    FOR    FINDIJVG    THE    ROOT    OF    ANY    MONOMIAL. 

Extract  the  root  of  the  numerical  coefficient,  and  divide  the 
exponent  of  each  literal  factor  by  the  number  which  marks  the  de- 
gree of  the  root. 

The  reason  for  this  rule  is  manifest,  since  extracting  a  root  is 
the  reverse  of  finding  a  power.    Thus,  the  second  power  of  5  a  6 

is  25  a^  b^ ;  consequently,  the  second  root  of  25  a^  b^  is  5  a^  b^ 
zzzba^b^  or  5 ab.  In  like  manner,  the  third  root  of  125 a^ b^ c^ 
is  o  1^  63  c. 


Jb?4  ROOTS    OF    MONOMIALS  XU 

Art.  117.  With  regard  to  the  signs  which  affec  tne  rocts  of 
monomials,  observe,  that 

Every  root  of  an  even  degree  may  have  either  the  sign  -\- 
or  — .  This  is  manifest  from  the  formation  of  powers.  Thus, 
the  fourth  power  of -(-a  is  -|-  a^,  and  the  fourth  power  of  — a 
is  also  -|-  a^.  Therefore  the  fourth  root  of  -(-  «^  is  either  +  « 
c.r  —  a.  Hence,  to  any  root  of  an  even  degree,  we  commonl) 
prefix  rh. 

But  roots  of  an  odd  degree  have  the  same  sign  as  the  power. 
The  third  power  of  +  a  is  -f-  a^  ;  whereas,  the  third  power  of 
—  a  is  —  a^;  -\-a  is  therefore  the  third  root  of  -|-  a^,  and  —  a, 
that  of  —  a^. 

It  has  already  been  stated  in  Art.  ^8,  that  the  second  root  of 
a  negative  quantity  is  imaginary.     The  same  is  the  case  with 

any  even  root  of  a  negative  quantity.     Thus,  ( —  16)^,    ( —  a)% 

( —  o)^,  or  the  equivalent  expressions,  i/ — 16,  w —  a,  i/ —  a 
are  imaginary  quantities ;  for  no  quantity  raised  to  a  power  of  an 
nven  degree,  can  produce  a  negative  quantity. 

1.  Find  the  second  root  of  a'^m^i^. 

2.  Find  the  second  root  of  64x^y^. 

3.  Find  the  third  root  of  343  a^p^  q^. 

4.  Find  the  third  root  of  —  729  a^  b^  c^^. 

5.  Find  the  fourth  root  of  16  aH^e  ^12. 

4g2 

6.  Find  the  second  root  of 


7,    Find  the  third  root  of 


8.  Find  the  fifth  root  of  — 

9.  Find  the  seventh  root  of 


25  3;4- 
27x9y3 

64^12- 
32  flio  615 


3125.r5yio 

2II87g'{621^c7 
16384>8,,35- 

The  preceding  examples,  as  well  as  what  was  said  in  Art 
10*>,  relative  to  the  powers  of  products,  show,  that  any  root  of 
a  product  will  be  the  product  of  the  roots,  to  the  same  degree,  oj 
ruch  of  the  factors  of  this  product.     Thus,  the  second  root  of 


XLl.  ROOTS    OF    MONOMIALS.  185 

«^ b^ c*  IS  ah  c2,  which  is  the  product  of  the  second  rvK).3  v  f  a^, 
6^  and  c**,  the  factors  of  a^  b^  c'*. 

In  like  manner,  if  any  numerical  quantit}  is  divided  into  fac- 
tors which  are  exact  powers  of  the  required  degree,  (and  this 
may  always  be  done,  when  the  number  itself  is  an  exact  power 
of  that  degree,)  we  may  extract  separately  the  roots  of  these  fac- 
tors, and  then  multiply  these  roots  together.  For  example,  1764 
=  36  .  49,  the  second  root  of  which  is  6  .  7  =  42. 

Art.  118.  From  the  preceding  mode  of  finding  the  roots  of 
literal  quantities,  it  follows,  that,  if  the  exponent  of  any  factor  is 
not  divisible  by  the  number  which  expresses  the  degree  of  the 
root,  the  division  can  be  expressed  only,  and  gives  rise  to  frac- 
tional exponents.  Thus,  the  second  root  of  a  is  a^;  the  third 
root  of  a  is  a* ;  the  fourth  root  of  a^  is  a*. 

The  expression  a*  indicates  either  the  fourth  root  of  a^^or 

the  third  power  of  a*  ;  for  the  third  power  of  a^  is  a^        =  a*. 

li\  like  manner,  a*  denotes  either  the  fourth  power  of  q^  or  the 
fifth  root  of  a^. 

The  radical  sign  may  be  used  to  indicate  a  root  of  any  degree, 
if  we  place  over  it  a  figure  denoting  the  degree  of  that  root. 

—  3  — 

Thus,  ^       denotes    the   second    root;    i/     ,  the  third  root; 
Y     ,  the  fourth  root,  and  so  on.     Hence, 

v/«"  =  «^. 

^a3  =  a^,  - 


^«6 


,1 


Observe,  that  ^a  is  the  same  as  j^a,  the  2  over  the  sign  be- 
ing generally  understood.  We  see,  therefore,  that  in  the  prece- 
ding equivalent  expressions,  the  nuTiber  over  the  radical  sign  is 
the  same  as  the  denominator,  and  the  exponent  under  the  sign  is 
tie  s^me  as  the  numerator,  of  the  fractional  exponent 
16* 


186  ROOIS    OF    POLYNOMIALS.  XLil 

Art.  110.  By  means  of  exponents  eith'rT  entire  or  fractional, 
any  quantity  may  be   expressed  in    a   greU  variety  of  forms 

Thus,  c?:=.c^.az=:a.a.a-=:.cu^.c^.(j^z=:a,cfi.c?.(j^,c^ 

-zz  a^  .  a^  .  a^  .  a^  .  a^  .  a^",  &.c.     Also,  ^a^  b^  z=z  a^   b^  =. 

a^  b^  ^=1  a  .  a^  .  b^  .  b^  z=z  a^  .  a^  .  a^  .  b^  .6^,  &c. 

Hence,  any  quantity  mav  be  separated  into  an  indefinite  num* 
ber  of  factors ;  the  only  restricuon  is,  that  the  sura  of  the  expo- 
nents of  those  factors,  which  are  alike  except  with  regard  to 
their  exponents,  shall  be  the  same  as  in  the  given  quantity.  In 
the  first  example  given  above,  the  sum  of  the  exponents  must  be 
uniformly  3;  in  the  second,  the  sum  of  the  exponents  of  a  must 
be  f ,  and  that  of  the  exponents  of  b  must  be  f . 

1.  Separate  a^  into  three  factors. 

2.  Separate  a*  into  seven  factors. 
*S.    Separate  a?  into  six  factors. 

4.  Separate  a  into  three  factors. 

2 

5.  Separate  d^  b  into  four  factors. 

6.  Separate  3  a^  into  six  factors. 

7.  Separate  35  into  three  factors. 

8.  Separate  10  into  seven  factors. 


SECTION   XLII. 

ROOTS    OF    POLTNOMIAXiS. 

Art.  ISO.   Let  it  be  required  to  find  the  second  root  ol  4  m* 
|-12»inf9ii2. 

Operation. 
4ywS+12mri  +  9nS(2m  +  3n.     Root. 
4n^ 

12mn  +  9n3(4m  +  3n. 
I2mn-i-9n^ 
0. 


tLII  ROOTS    OF    POLyNOMIALli.  IfcO 

By  recurrii.^  to  the  second  power  of  a  -{-  6,  which  is  g^  f-SaA 
•}-  62,  we  see  that  4  m^  corresponds  to  a\  We  therefore  take  the 
second  root  of  4  m^,  which  is  2»i,  and  place  it  at  the  right,  as  the 
first  terra  of  the  root  sought,  and  subtract  its  second  power  from 
the  given  quantity.     The  remainder,  12  wt  w  -|-  9  w^,  answers  to 

1  ab-\-h^y  or  {^  a -\-h)h.  Dividing  the  first  term  of  this  remain- 
der by  4  m,  corresponding  to  2  a,  we  have  3  n  for  the  second  term 
of  the  root,  which  we  annex  to  the  2  m  in  the  root,  and  also  to 
the  divisor.  The  divisor  thus  increased,  becomes  4  m  +  3n,.=: 
Za-\-h.  We  then  multiply  4  m  +  3  n  by  3  w,  =  6,  and  we  have 
12  m  w  4~  9  '^^j  which  subtracted  from  the  dividend,  leaves  no  re- 
mainder.     Hence,  the  second  root  of  4  m^  -j-  12  m  »  +  9  w^  is 

2  m  +  3  w,  or  —  2  m  —  3  w ;  or  rather,  i  2  m  zh  3  w. 

The  double  sign  may  be  omitted,  until  the  operation  is  com- 
pleted, and  then  all  the  signs  of  the  root  may  be  changed,  if  both 
roots  are  required. 

When  there  are  more  than  three  terms  in  the  power,  the  sec- 
ond root  will  contain  more  than  two  terms.  But  the  mode  of 
proceeding  will  be  almost  the  same  as  that  for  finding  the  second 
roGls  of  numbers.  We  form  a  second  dividend,  in  the  same  man- 
ner as  the  first  was  formed,  and  for  a  divisor  double  the  whole 
root  found.  The  division  will  give  the  third  term  of  the  root. 
The  process  is  manifest  from  the  formula,  \a-\-b-\-c-\-d,6i,c.)^ 
=  a2-|-2a6  +  62^2  (a  +  6)c  +  c2  +  2(a  +  6  +  c)c?+d2, 
&c.,  in  which  a,  6,  c,  &,c.  represent  the  teinis  of  the  root. 

The  following  example  will  serve  to  illustrate  the  process. 

Required  the  second  root  of  a'*  +  6  a^  z  +1 1  a^  a;^  -|-  6  a  x^  -|-  a^ 

Operation. 

a4 

Qa^T+nd^x^  +  Qax^  +  x^'ia^+^ax. 

6q3z+    9q^^^ 

2a2  2;2_|_6a2;3_|_2;4  (2a2-f6  gx  +  x^ 

2a2a:S-|-6az3-t-2;^ 

a 

The  root  required  is  a^^Sax-^xK 


188  ROOTS    OF    POLYNOMIALS.  XLIl 

From  the  preceding  analysis  we  derive  the  following 

RULE    FOR    EXTRACTING    THE    SECOND    ROOT    OF    A.    POLYNOMIAL. 

1.  Arrange  the  quantity  according  to  the  powers  of  somt 
letter. 

2.  Find  the  root  of  the  first  term^  and  place  it  as  the  first  term 
of  the  root  sought,  subtract  the  second  power  of  this  term  from  the 
given  polynomial,  and  call  the  remainder  the  first  dividend. 

3.  lyouhle  the  term  of  the  root  found,  for  a  divisor,  by  which 
divide  the  first  term  of  the  dividend,  and  place  the  quotient,  with 
its  proper  sign,  as  the  second  term  of  the  root,  also  at  the  right 
of  the  divisor.  Multiply  the  divisor,  with  the  term  annexed,  by 
the  second  term  of  the  root,  and  subtract  the  product  from  the 
dividend. 

4.  The  remainder  will  form  a  new  dividend,  which  is  to  be 
divided  by  twice  the  whole  root  found,  and  the  quotient  is  to  be 
placed  as  the  next  term  of  the  root,  also  at  the  right  of  the  divi' 
sor.  Multiply  the  divisor,  with  the  term  last  annexed,  by  the  last 
term  of  the  root,  and  subtract  the  product  from  the  last  dividend. 

5.  The  remainder  will  form  a  new  dividend,  with  which  pro- 
ceed as  before ;  and  thus  continue,  until  all  the  terms  of  the  root 
are  found 

Remark  1.  Each  of  the  remainders  must  be  arranged  in  the 
same  order  as  the  given  polynomial  was  first  arranged. 

If  the  given  quantity  contains  no  fractions,  and  a  dividend  oc* 
curs,  the  first  term  of  which  does  not  contain  all  the  letters  o\ 
the  first  term  of  the  divisor,  or  which  contains  any  one  of  then» 
with  a  less  exponent  than  it  has  in  that  term  of  the  divisor,  wp 
may  be  assured  that  the  given  quantity  is  not  an  exact  second 
power,  and,  therefore,  does  not  admit  of  an  exact  root. 

Remark  2.  In  dividing  we  merely  divide  the  first  term  of  the 
dividend  by  the  first  term  of  the  divisor ;  and,  since  double  the 
first,  the  first  two,  the  first  three,  &c.  terms  of  the  root,  will  have 
the  first  terms  alike,  it  is  manifest  that  the  successive  divisors 
will  have  their  first  terms  the  same. 

Find  the  second  roots  of  the  fVlowing  quantities 


KlJl.  ROOTS    OF    POLYNOMIALS.  189 

1.  3iz^+9x*  +  20x-{-\2x^-{-25. 

2.  a4_|_54«252_|_i2a36+108«63  +  8164. 

3.  ]03;4_l0z3_i2x5  +  5a;2-|-92;6_2x+l, 

4.  9a4  — 20a63_12a36  +  34a262  +  2564. 

5.  4x4  +  8az3  +  4a2x2+1662x2_|_i6a622;^1064. 

6.  x^  +  4x5-{-2x^+9x^  —  4x-{-i. 

7.  4z4-f  6  a:3_j_  5^x2.^15  2;  _|.  25. 

Art.  121.  The  rule  for  extracting  the  third  roots  of  numbers, 
might,  with  slight  modifications,  be  applied  to  the  extraction  of 
the  third  roots  of  algebraic  polynomials.  But  it  is  generally  the 
most  convenient  to  use  the  rule,  derived  from  the  binomial  theo- 
rem, for  the  extraction  of  roots  to  any  degree.  This  rule,  ap- 
plied to  literal  quantities,  will,  as  is  evident  from  the  formula  for 
the  wth  power  of  x  -|-  «,  be  as  follows. 

BULE    FOR    EXTRACTING    ANY    ROOT    OF    A    POLYNOMIAL. 

1.  Arrange  the  quantity  according  to  the  powers  of  some 
letter. 

2.  Find  the  mth  root  of  the  first  term^  place  it  as  the  first  term 
of  the  root  sought ^  and  subtract  the  mth  power  of  it  from  the  poly" 
nomial. 

3.  The  remainder  will  form  a  dividend^  which  is  to  he  divided 
by  m  times  the  (m  —  \)th  power  of  the  term  of  the  root  found, 
and  the  quotient  is  to  be  placed  as  the  second  term  of  the  root. 

4.  Raise  the  whole  root  to  the  mth  poioer,  and  subtract  the  rcr 
suit  from  the  given  polynomial. 

5.  The  remainder  will  form  a  new  dividend^  which  is  to  be  di- 
vided by  m  times  the  (m  —  \)th  power  of  the  whole  root  already 
founds  and  the  quotient  placed  as  the  third  term  of  the  root. 

6.  Raise  the  whole  root  to  the  mth  power,  subtract  the  result 
from  the  given  polynomial,  and  with  the  remainder  proceed  as 
before;  and  thus  continue  until  all  the  terms  of  the  root  are 
found. 

Remark.  It  is  manifest  that  the  first  term  of  each  successive 
divisor  will  be  the  same ;  and,  since  we  always  divide  the  first 
term  of  the  dividend  by  the  first  of  the  divisor,  it  is  sufficient  to 
find  the  first  term  of  the  first  divisor  and  use  that  throughout ; 


190  ROOTS    OF    POLYNOMIALS.  XLU 

and^  in  subtracting,  only  one  term  of  the  remainder  reeds  to  be 
brought  down,  viz  •  that  which  contains  the  highest  power  of  the 
letter  according  to  which  the  given  quantity  was  arranged. 

As  an  example,  let  it  be  required  to  extract  the  thir^  root  of 
8  x3  +  60  x2  y  -|-  150  xi/^-{- 125  y3. 

Ojperation. 
82;3-f  60  z2y_^  150  xy2_j_  125^3  (2^_j_5y,     Root. 

8x3 

60xQy  (12a:2.     Divisor. 

The  index  m  of  the  general  formula,  when  applied  to  this 
question,  is  3 ;  and,  after  having  arranged  the  quantity  according 
to  the  powers  of  x,  we  find  the  third  root  of  Sx^,  which  is  2x; 
subtracting  the  third  power  of  2  x,  we  have,  for  the  first  term  of 
the  remainder,  60x2y,  which  we  divide  by  12x2,  ■=  three  times 
the  second  power  of  2  x.  The  quotient  is  5  y,  which  we  put  as 
the  second  term  of  the  root,  and  raise  2  x  -|-  5  y  to  the  third 
power  *  The  result  is  the  same  as  the  given  quantity,  and,  when 
subtracted,  leaves  no  remainder.  Therefore,  2x-|-5y  is  the 
root  sought. 

As  a  second  example,  we  shall  trace  the  operations  for  extract- 
ing a  root  consisting  of  three  terms. 

Let  it  be  required  to  extract  the  fifth  root  of  x^^  —  10  x^  a  -|- 
45  xS  «2  _  120  x7  a3  _}_  210  x^  a^  —  252  x^  aS  _[_  210  x^  a^  — ' 
120  x3  a?  -f  45  x2  a8  _  1 0  X  a9  _^  a^o. 

The  quantity  being  arranged  according  to  the  powers  of  x,  we 
find  the  5th  root  of  x^^,  which  is  x^,  and  subtract  the  5th  power 
of  this  root  from  the  given  quantity.  The  first  term  of  the  re- 
mainder is  —  10  x9  a.  This  term  we  divide  by  five  times  the  4th 
power  of  x^,  which  is  5  x^.  .The  quotient,  — 2 ax,  we  place  as 
the  second  term  of  the  root,  and  raise  r^  —  2  a  x  to  the  5th 
power.  The  5th  power  of  x^  —  2  a  x  is  x^*^  —  10  x^  a  -|-  40  x®  a^ 
—  80  x^  fl3  _|_  80  x6  a'*  —  32x5a5^  which  subtracted  from  the  given 

*  Let  the  learner  use  the  binomial  theorem  for  finding  the  powers  of  ai  y 
quantity  conrfsing  of  more  than  one  term. 


X.LIIL       SIMPLIFICATION    OF    IRRATIONAL    QUANTITIES.  191 

quantity,  gives  a  remainder  the  first  term  of  which  is  5x^a^ 
This  tp«m  being  divided  by  5  x^,  the  first  term  of  five  times  the 
4th  power  of  x'^  —  2 ax,  gives  for  a  quotient  a^,  which  we  place 
as  the  third  term  of  the  root.  We  then  raise  x^  —  2ax-\-a^  to 
the  5th  power,  and  it  produces  the  whole  of  the  given  quantity 
Hence,  x^  —  'Hax-j-a^  is  the  root  sought. 

1.  Findthe3drootof27a3-|-81az2_|.81a2z  +  27  23. 

2.  Find  the  4th  root  of  16  x^a  +  1000  7^  cfi -\- 600  x^^  a^  -\^ 
160  29a2  +  625a«. 

3.  Find  the  4th  root  of  625  c^  —  1 000  c^  y  z  +  600  c*  y^  z^  — 
160  c2y3  23_|_  16^4^4. 

4.  Find  the  5th  root  of  32  a^^ — 80  a^  h^  +  80  a^  je  _  40  a4  59 
-J.  10  a2  612  — 615. 

5.  Find  the  6th  root  of  729  x^  -f-  2916  x^  y  +  4860  x^y^-\- 
4320  x3  y3  _|_  2160  x2  y^  +  576  x  yS  _|_  54  ^6. 


SECTION    XLIU. 

SIMPLIFICATION    OF    IRRATIONAL    OR    RADICAL    QUANTITIES. 

Art.  IS^.  When  a  quantity  is  not  an  exact  power  of  the  de- 
gree required,  its  root  cannot  be  found  exactly.  In  such  a  case, 
the  root  is  commonly  expressed  either  by  a  radical  sign  or  by  a 
fractional  exponent.  Expressions  indicating  roots  which  cannot 
be  accurately  obtained,  are  called,  as  has  already  been  stated^  ir- 
rational or  incommensurable  quantities.    They  are  also  sometimes 

called  surds  or  simply  radical  quantities.     Thus,  ^2  or  2*  and 

3—1 

/4  or  4^  are  irrational  quantities. 
In  like  manner,  we  are  obliged  to  express  the  second  root  of  o 

by  a  sign,  thus  y/«  or  a^ ;  although,  perhaps,  when  a  has  been 
replaced  by  its  numerical  value,  the  root  may  be  exactly  found. 
Algebraically  considered,  however,  such  expressions  are  in  th* 
condition  of  irrational  quantities. 


192  SIMPLIFICATION    OF  X  LTIl 

Expressions  of  this  kind  may,  in  many  cases,  be  simplified. 
The  root  of  a  product,  as  was  shown  in  Art.  117,  is  formed  b) 
multiplying  together  the  roots  of  all  the  factors  of  that  product 
Hence,  we  may  take  the  roots  of  such  factors  as  are  exact  pow- 
ers, and  indicate  the  roots  of  the  other  factors^  leaving  these  roots 
to  hi  approximated  afterwards  if  necessary. 

Let  it  be  required,  for  example,  to  find  the  second  root  of 
1 92  0.2  6-3  t.  The  root  is  indicated*  thus,  ^192  a^b^c.  But 
192«263c  =r  64  .  3a262^c  or  64a^b^.Sbc.  Now  the  first 
three  factors,  64,  a^  and  6^,  are  second  powers^  we  may,  there- 
fore, take  the  roots  of  these  and  place  their  product  as  a  coeffi- 
cient to  the  expression  indicating  the  root  of  3  6  c.     We  have 

then  1/192  a^b^  c  =z8ab  i/3  b  c.  It  only  remains  now  to  ap- 
proximate the  root  of  3  6  c,  the  value  of  the  letters  supposed  to 
be  known,  and  multiply  the  result  by  8  a  6. 

In  separating  an  irrational  quantity  into  factors  for  the  pur- 
pose of  simplifying,  the  learner  has  merely  to  find  the  greatest 
numerical  factor  that  is  an  exact  power,  and  the  greatest  expo- 
nent of  each  literal  factor,  not  exceeding  its  given  exponent,  that 
is  divisible  by  the  number  which  marks  the  degree  of  the  root. 


1. 

Simplify  ^125a3  65. 

2. 

Simplify  (80«6c4)*. 

3 

3. 

Simplify  |/108a9  66  c2. 

4. 

Simplify  ^45  «  62  c2. 
3 

5.    Simplify  ^320  a3  6  —  64  a^  bK 

The  greatest  numerical  factor  in  this  quantity  that  is  a  third 
power  is  64,  and  the  greatest  literal  factor  that  is  a  third  power 
is  a3.  HenQe,  320  a^b  —  Qi  a^  6^  =  64  a^  (5  6  —  a^  63).  Tak- 
ing the  root  of  64  a^y  and  indicating  that  of  5  6  —  a®  53^  ^q  h^ve 

^320«3  6_64a5  63  _  4  ^  ^5b  —  a^b'^,  or  4 a  (56— a2 63)* 


G.    Simplify  ^24«4__8a36. 
7.    Simplify  (2  a^  b^  +  a^b  c)K 


JtLIII  IRRATIONAL    QUANTITIES.  193 


8.    Simplify  ^a^  -\-  d^  b^. 


9.    Simplify  y/TeSa^  — 256aX 


10.    Simplify  y/3456 a^b—\ 728. 

If  the  quantity  which  is  under  the  radical  sign,  or  which  is  en. 
closed  in  a  parenthesis  with  a  fractional  exponent,  is  a  fraction, 
the  expression  may  be  simplified  in  the  following  manner,  viz : 

Multiply  both  terms  of  the  fraction  bp  such  a  quantity y  as  will 
render  the  denominator  an  exact  power  of  the  requisite  degree^ 
then  take  the  roots  of  the  denominator  and  of  such  factors  of  the 
numirator  as  are  exact  powers. 

Remark.  This  preparation  of  the  fraction  is  rarely  advisable, 
except  when  the  denominator  is  a  monomial. 


Thus,   1  /^  =  t    /6«^*  =  .    /'^'_     eab  = 
'   V      8b        Y      1662        \/     1663°"" 

h.^^'-    _  

la  like  manner,  I  Vl^^  =  .  V 12^^  = 
K      94        V-      27  63 


V^. 


2^L..12a«63=i-^^12„«6.. 


11   Simplify  y  |f 

12.   S.mpl.f,  {-^^f. 

U.   Simplify  ^g. 

14.  Simplify  (ll)*. 

15.  Simplify  y/^~. 

16.  Simplify  I  yK^'-l?4^. 

17 


194  IRRATIONAL    QUANTITIES  XLIV 

17     Simplify  (-500^4_250^7r) 


18.    Simplify  ^  27^5^^ 


320  a3  6  4-  640  a^ 
54  c3  m^  ' 


Art.  1S3.  As  we  can  extract  the  root  of  any  factor  and  place 
it  as  a  factor  before  the  radical  sign  ;.so,  if  we  would  put  under 
the  sign  any  factor  standing  before  it,  we  must  raise  that  facto* 
♦o  a  power  of  the  same  degree  as  the  radical. 

Thus  a  b  ^c   =  \J(j^  b^  c ;  and  |  ya  m  =.  \/     — wr-  . 
Reduce  the  following  quantities  entirely  to  a  radical  form. 

1.  3azy/6r  '       5.   2a (36)^. 

2.  §v/^^-  6.   S{xy)^, 


3. 


4.    («  + 6)^/3^.  8.   ||(m^)i 


SECTION    XLIV. 

OPERATIONS    ON     IRRATIONAL    QUANTITIES    WITH    FRACTIONALi    EX- 
PONENTS. 

Art.  1^4:.  In  general,  operations  are  performed  on  quantitiei 
with  fractional  exponents,  in  the  same  manner  as  if  the  exponents 
rere  entire. 

Add  4a^  and3a^. 

The  sum  is  4  a^  +  3  a^  =  7  a^. 
Add  a  b^  and  c  b^. 

The  sum  is  ab^  +  c  6^  =  (a-f-c)6^. 


XLIV  WITH    FRACTIONAL    EXPONENTS.  I9& 

From  9  z*  subtract  4  z*. 

The  difference  is  9  z^  —  4  z^  =  5  z^. 
From  3  a  z^  y^  subtract  2  6  z-  y^. 

The  difference  is  3az^y*  —  26z^y*=(3a— 26)z^^^ 
Add  3  .  8^  and  5  .  18^. 
The  sum  indicated  is  3  .  8^  +  5  .  18^.  But  these  terms 
may  be  simplified  and  the  expression  reduced.  For,  3.8^  = 
').  4^.  2^  =3.  2.  2^  =  6.2^;  and  5.18^  =  5.9^.2^  = 
5 .  3  .  2*  =  15  .  2^  Hence,  3 .  8^'+  5  .  18^  =  6  .  2^  +  15 .  2^ 
=  21(2)^. 

In  a  similar  manner,  ( 192  63) *  +  (24  c^)^  =  4  6 .  3*  +  2  c  .  3^ 

=:(46  +  2c)3t 

Add  (^\)^  and  ii)^. 

The  sum  expressed  is  (^)*  +  (i)^.     But  (^)^  =  (f f)^ 
=  (^^.6)^  =  1. 6*;  and(#  =  (^e)^  =  (^.6)^=  J.6^ 

Hence  (^)^  +  a)^  =  f.6^  +  i.6*=:A.6^  +  A.6^ 
=  /f  (^)  •     'T^®  result  therefore  in  its  simplest  form  is  -^^  (6)* 

We  deduce  therefore  the  following 

RULE     FOR     THE     ADDITION     AND     SUBTRACTION    OF     IRRATIONAi^ 
Q.UANTITIKS. 

Express  the  addition  or  subtraction  as  usual  by  signs ^  sim 
plify  the  terms  if  possible  ^  and  reduce  similar  terms. 

Remark.  Irrational  quantities,  indicated  by  means  of  frac- 
tional exponents,  are  similar,  when  the  factors  having  fractional 
exponents  are  alike  in  all,  and  have  severally  the  same  expo- 
nents. 

Multiply  a^  bv  a^. 
This  is  performed  by  adding  the  exponents.     Thus,  a^    a» 


196  IRRATIONAL    QUANTITIES  XLIV 

Multiply  3a^6^  by  5a^6^. 

The  product  is  15  J^^  b^^^  =  lo  J  6* 
Multiply  2  a^  by  3  a^. 

Reducing  the  exponents  to  a  common  denominator,  wt 
ha\  e  2a^  =  2  a^^,  and  3  a^  =  3  a^^ ;  therefore,  2  a^  .  3  a*  = 
2a^  .3a^^=:6«^t 

Divide  a^  by  a^.  ' 

This 'is  performed  by  subtracting  the  exponent  of  the  latter 

J 

from  that  of  the  former.     Thus,  —  =  a^    ^  =  a' 


Divide  15  a^  b^  by  3a^  6^ 

1  he  quotient  is  ——-z=ba^    •*. 

3«^6* 

Divide  3a^6^  by  5a^  fti 

Reducing  the  exponents  of  the  similar  factors  to  a  common 

3a*  6^       sJb^       sJb^ 
denominator,  we  have  — - — -  =  — - — —  = . 

5«*6*       5a^b^^  ^ 

2 

Find  the  second  power  of  3  a^. 

This  is  performed  by  raising  the  coefficient  to  the  required 
power,  and  multiplying  the  exponent  by  the  index  of  the  power. 

Thus,    (3  a^)  ^  =  9  a^. 

In  like  manner,  (3  a^  b^)  *  =  81  J  b^. 

Conversely,  the  root  of  an  irrational  quantity  is  found,  by  taking 
or  expressing  the  root  of  the  numerical  coefficient,  and  dividing 
the  exponents  of  the  other  factors  by  the  number  which  marks 
♦he  degree  of  the  root. 

T  ms,  the  second  root  of  of  is  a^,  and  the  fourth  root  of  a" 


XLIV  vvnn  fractional  exponent**.  197 

is  a^.     Also  the  third  root  of  21  a^  b^   is  3a*  b^^  ;   and  the 

fifth  root  of  7  J  6*  is  7^  a"^  6^^. 

From  what  precedes,  we  see  that  the  following  operations,  viz  : 
multiplication,  division,  finding  powers,  ahd  extracting  roots ^  are 
performed  upon  quantities  with  fractional  exponents,  in  the  same 
manner  as  if  the  exponents  were  whole  numbers. 

Art.  Id^.  In  the  multiplication  of  irrational  quantities,  we 
assumed  that  their  fractional  exponents  might  be  reduced  to 
equivalent  ones  having  a  common  denominator,  without  chang- 
ing the  value  of  the  quantities.  This,  however,  may  be.easily 
proved. 

For,  multiplying  the  numerator  of  the  exponent  c  f  any  quan- 
tity, raises  that  quantity  to  a  power,  and  multiplying  the  denom- 
inator, divides  the  exponent,  and  therefore  extracts  the  root. 
Consequently,  when  both  terms  of  a  fractional  exponent  are  mul- 
tiplied by  the  same  number,  which  is  done  in  reducing  to  a  com- 
mon denominator,  the  quantity  to  which  the  exponent  belongs,  is 
raised  to  a  power  of  a  certain  degree,  and  then  the  root  of  the 
result  is  extracted  to  the  same  degree ;  or  the  reverse.  The 
value  of  the  quantity,  therefore,  remains  unchanged. 

Accordingly,  the  exponents  of  all  the  factors  in  any  product, 
may  be  reduced  to  fi  actions  having   a  common  denominator. 

Thus,  2a^  b^  —  2^  a^  b^  =  64*  «*  b^  z=  (64  a^  b^)^. 

Moreover,  it  is  manifest  that  the  fractional  exponents  may 
be  reduced  to  decimals,  and  that  the  value  of  the  result  will  be 
either  exactly  or  approximately  the  same  as  that  of  the  given 
quantity. 

For  example,  a*  =z  a°'^^,  and  a®  =  aO-875^    If  \i  ^q^q  required 

then  to  multiply  a*  by  a^,  we  should  have  a*  .  a^  ==  a^  25  .  afwi 

_-  ^0-25 +  0-875  — -  al*125^ 

1.    Add  {27  a*  x)^  and  (Sa^x)^. 

2     Add  (128)^  and  (72)^. 

3.    Add  (135)^  and  (40)i 
17* 


19S  IRRATIONAL    QUANTITIES  XLl*/ 

4.  Add  (12)^,  2  (27)^  and  3  (75)^. 

5.  Add2(8)*— 7(l8)*and5(72)^  —  (5()^ 

6.  Add- 7  (54)^,  3  (.16)^  and  (2)^  —  5(128)^ 

7.  Add  (4  a2  6)^,  3  a  6^,  (27  a^  6)^  and  ( 125  aS  ft) 

8.  Add  3  (^V)^  and  4  (f)^. 

10  .Add  {9abf,  (c^ab)^,  (^^Y  and  (xy)*      * 

11.  From  (18)^  subtract  8^ 

12.  From  ( 108 a x'^)^  subtract  (48 ax^)^. 

13.  From  (432  a^  b)^  subtract  (16  a?  b)^. 

14.  From  ( 192  a^  ftsji  subtract  (24  a^  b^)t 

15.  From  f  (f  )^  subtract  f  (^)^. 

16.  From  5  (20)^  subtract  3  (45)^. 

17.  From  (16  a  6)^  —  (343  m)^  subtract  (9  a  6)*  — 

(1000c3m)t 

^^-  ^^^™  (l92  j    -(l35)^"^*'^^K-27-)    -(W 

19.  Multiply  7  a*  6^  by  3  a^  6^. 

20.  Multiply  2  a  6  c  by  5  a^  6^  c^. 
21  Multiply  m  a^  c*  by  3  m  a^  A 

22.  Multiply  25  x^  y  by  3  x^  y*. 

23.  Multiply  10  (108)*  by  5  (4)^. 

The  product  is  50  (108)*  (4)*  =  50  (432)*  =  60  (216   2)^ 
=  50   6 .  2*  =  300  (2)^. 

24.  Multiply  5  .  5*  by  .3 .  8^,  and  simplify 


XLIV.  WITH    FRACTIONAL    EXPONENTS.  199 

25    Multiply  2  .  sHy  3  4^ 
The  product  is  6.  3*,  4^  =  6   3^.4^  =  6.27*    16*  = 

6(432)* 

26.  What  is  the  product  of  4,  2  (3)*  and  72*? 

27.  What  is  the  product  of  5  (3)*,  7  (f  )^  and  2*t 

i.  1 

28.  Multiply  (T  by  o". 

m  p 

29.  Multiply  3  a"  by  5  a^ . 

30.  Multiply  together  2^,  3*  and  5* 

31.  Multiply  (a +  6)*  by  (a +  6)i 

32.  Multiply  3  (c  —  d^  by  4  a  (c  —  rf)*. 

33.  Multiply  4 aS  6  (x—y)*  by  3(0  +  6)*  (x— y)*. 

*    34.  Multiply5(m  +  n)*(c  — rf)*by7(i»  +  »:*(c  — rf)^. 

a5.  Multiply  3  +  5*  by  3  —  5*. 

36.  Multiply  7  +  2  (6)^  by  9  —  5  (6)* 

37.  Multiply  9  +  2(10)*  by  9  — 2(10)*. 

38.  Divide  a^  by  a*. 

39.  Divide  a6*c  by  a^  6^  c*. 

40.  Divide  6  a^^  6^  c^  by  3  a*  6  c^, 

41.  Divide  3  a^  fe^  c^  by  4  a*  6^  c* 

42.  Divide  10(108)*  by  5  (4)^. 

43.  Divide  10  (27)*  by  2  (3)*. 

44.  Divide  8  (512)^  by  4(2)* 

45.  Divide  a"*  by  a". 

46  Divide  3  a"  6"  by  4  0^6'. 

47  Divide  ^(§)*  by  ^  ( J)* 


200  IRRATIONAL    QUANTITIES  XIJV 

48.  Divide  (a  — 6)^  by  (a  — 6)i. 

49.  Divide  12(x  — y)^by  4a(x  — y)i 

50.  Divide  13  {a  +  b)^  (c  —  d)^  by  39  m  (a  -f-  b)^  (c  —  rf)* 

51.  Find  the  2d  power  of  2  a^  b^. 

52.  Find  the  3d  power  o{5a^b^  A 
2a^6^ 


53.    Find  the  3d  power  of 


Sx^f/ 


54.  Find  the  4th  power  of  ^  .  3^. 

2  .  3^  fli  6  c2 

55.  Find  the  5th  power  of  — '- — , 

50.  Find  the.  3d  power  of  {a-\-b)^. 

57.  Find  the  3d  power  of  3  (x — y)^. 

58.  Find  the  5th  power  of  2  (x2  — y9)i  (b  —  r)*. 

59.  Find  the  »»th  power  of  a^  b^  c. 

60.  Find  the  7«th  power  of  a*  6*  c  «?. 

61.  Find  the  3d  power  of  ^-^^^-i^. 

62.  Find  the  2d  power  of  ?ii±y)j£Zlf(l^. 

5(m  +  n)*(x— ^)t 

63.  Extract  the  2d  root  of  a^  b^. 

64.  Extract  the  3d  root  of  27  a^  b^. 

65.  Extract  the  3d  root  of  2  a^  b^, 

66.  Extract  the  4th  root  of  3  a^  b^  c, 

67.  Extract  the  wth  root  of  10  a*  x  y^. 
68'  Extract  the  with  root  of  6  a'  6*. 


XLV  WITH    THE    RADICAL    SIGN.  20 1 

69  Extract  the  2d  root  of  {a-\-b)^. 

70  Extract  the  2d  root  of  16  (a  —  6)^c  —  rf)^. 

71.  Extract  the  3d  root  of  Sa^{m  —  n)^  (c  —  d)^. 

72.  Extract  the  3d  root  of  ^  (-' - y)U-' +  d)\ 

5b{m  +  n}^ 


SECTION    XLV. 

OPERAriONS  UPON    IRRATIONAL  QUANTITIES  WITH    RADICAL  :>IGNt 

Art.  130.  Although  radical  signs  may  be  wholly  dispensed 
with,  and  fractional  exponents  used  instead  of  them,  yet,  as  these 
signs  occur  in  almost  all  mathematical  treatises,  and  are  some- 
times very  convenient,  we  shall  show  how  to  perform  the  various 
operations  on  quantities  affected  with  them. 

Irrational  quantities  affected  with  radical  signs,  are  commonly 
called  radical  quantities,  the  mode  of  simplifying  which  has 
already  been  shown. 

The  addition  and  subtraction  of  radical  quantities  are,  it  is 
manifest,  performed  in  the  same  manner  as  when  fractional  ex- 
ponents are  used.  We  observe,  however,  that  radical  quantities 
are  said  to  be  similar,  when  they  are  of  the  same  degree  and 
nave  the  quantities  under  the  sign  in  all  respects  similar. 

3  3  — 

Let  it  be  required  to  add  together  ^192  and  y/24.  The  sum 
expressed  is  y^l92  +  ^24.  But  ^i92  z=z  ^64.3  =  4  ^S, 
and  ^24  =  ^873  =  2^3;  hence,  ^192 +^24  =  4^3 -f- 
2^3r=6y^3. 

In  like  manner,  the  sum  of  m  \^a^  h^  c  and  6^  „  */q3  ^  ^ 
^ ah^ m -\- ah^ n)  \^a c,  or  ah^{m-\- n)  y^a c. 


202  IRRATIONAL    QUANTITIES  X  L V 

8 3  — 

Subtract  ^108  from  9^4.       The    difference   expiessed   ia 

3  —  8  8  —  3  — 

9^4  — ^108,    which  simplified   becomes  9^4  — 3i/4  = 
6^4. 

In   like   manner,  i/o^  6   subtracted  from   3  m  wc^  b,  leaves' 

4   -  4  —  4  - 

dmc \/h  —  a  \^h  =.  (3 m  c  —  a)  ^b. 

Art  137.  Rules  for  other  operations  on  radicals,  may  be 
easily  deduced  from  the  modes  given  in  the  preceding  section, 
for  performing  corresponding  operations  on  irrational  quantities 
with  fractional  exponents. 

2%c  exponents  of  the  quantities  under  the  radical  sign  and  the 
index  over  that  sign,  may  both  be  multiplied  or  divided  by  the 
same  number  without  affecting  the  value  of  the  expression. 

2 8  15  10  

For  example,  ^a^  or  i/a^  z=  a^  =.  dP^  =  ^a^^,  which  might 
have  been  obtained  by  multiplying  the  2  and  3  of  the  expression 

2  — 

^a3  both  by  5. 


In  like  manner,  i/a^  b  =  i/a^  b^. 

10 15  3  2 

Again,  ^a^^  =  d^^  =z  a^  z=z  \Jd^  or  ^a^,  which  is  obtained 

10 

by  dividing  the  10  and  15  of  the  expression  ^a^^  both  by  5. 
In  a  similar  manner,  i/a^  6^  ■=.  wa^b. 
Art.  128.    Hence,  two  or  more  radical  expressions  may  be 

—  3 

made  to  have  the  same  index  over  the  sign.    Thus,  ^a^  and  ^b^ 

6 —  6  —  4  6  

are  respectively  the  same  as  wa^  and  i/6^ ;  also i/a^  b  and  Wx  y^ 

12  12 

are  respectively  the  same  as  i/a^i^  and  ^x^y^. 

This  process  is  evidently  the  same  as  reducing  fractional  expo- 
Qents  to  a  common  denominator,  the  indices  over  the  sign  being 
considered  as  denominators,  and  the  exponents  of  the  quantities 
under  the  sign,  as  numerators. 

The  common  index  will  therefore  be  the  product  of  all  tho 
indices  over  the  sign,  or  their  least  coramor  multiple. 


0  a^  b^ 


XLV  WITH    THE    RADICAL    SIGN.  203 

Art.  139.    Multiply  ^a  by  ^6. 

The  product  is  Wa  b ;  for  ^a  =.  a^,  and  ^b  =  b^ ;  there* 

fore,  ^a  .y/b  =  a^  6*  =  (a  6)^  =  ^^. 

Multiply  2  y/'^  by  3  y^P! 
We  first  render  the  indices  over  the  sign  alike ;  we  then  have 

2  j/^  .  3  v/P  =  2^^  .  3  y/'fts  =  6a^^  6 *  =  6  (a^o  69)^^ 
~  6  ya^K 

Divide  Q^ab  by  3^/a.     The  quotient  expressed  is     V^ 

3^a 

Rut  6y/^  =6a^  6^,  and  3^a  =  3a^;  therefore  ^V^"_^    == 

=r26^=:2y/6: 
3a^ 

—  3  — 

Divide  3i/a  by  5i/6  Making  the  indices  alike  and  then 
dividing,  we  have  -J- —  z=i  — 5- ■=.  -  .  —  =z:L(_A^  — 

Hence,  we  have  the  following 

RULE    FOR    THE   MULTIPLICATION    AND    DinsiOW    OF    RADICALS. 

Make  the  indices  over  the  radical  sign  alikcy  if  they  are  not 
Stj ;  then  multiply  or  divide  one  coefficient  by  the  other ;  also  take 
the  product  or  quotient  of  the  quantities  under  the  radical  sign, 
placing  the  latter  result  under  the  common  sign,  which  is  to  be 
preceded  by  the  product  or  quotient  of  the  coefficients. 

Art.  130.    Find  the  fifth  power  of  ^^^T. 
Since  2  ^^^  =2d^  b^,  we  have  (2)/^Tf  =  (2  a^  6*)* 
=  25  .  J^^  b^^^  =  22a^  b^  =  32  (a^^b^)^  =  32^aiop 


204  IRRATIONAL    QUANTITIES  XLV 

This  result  might  have  been  obtained  from  2  i/a^  b    by  rais« 

hig  the  coefficient  2  to  the  fifth  power,  and  multiplying  the  expo* 

nents  o/*the  quantities  under  the  radical  sign  by  5. 

9  — 
Raise   3  wa^  to  the  third  power. 

Since   S^^  =  sJ,  we  have  (S^a^f  =  (3  J)^  i=  32»  X 

J^  =  27  a^  =  27  ^^.  This  result  is  obtained  from  S^^, 
by  raising  the  coefficient  3  to  the  third  power,  and  dividing  the 
index  9  by  3.     Hence  we  have  the  following 

RULE    FOR    RAISIIVG    A    RADICAL    TO    AKTY    POTVER. 

Raise  the  coefficient  to  the  power  required^  and  either  raise  the 
quantity  under  the  radical  sign  to  the  same  power ,  or  divide  the 
index  over  it  by  the  number  expressing  the  degree  of  the  power. 

Art.  131.  Since  extracting  a  root  is  the  reverse  of  finding  a 
power,  we  have  the  following 

RULE    FOR    EXTRACTING    ANY    ROOT    OF    A    RADICAl,. 

Extract  the  root  of  the  coefficient^  and  either  extract  the  root 
of  the  quantity  under  the  radical  sign,  or  multiply  the  index  over 
it  by  the  number  expressing  the  degree  of  the  root. 

Thus,  the  third  root  of  8  ya^  is  2  ya^ ;  and  the  fourth  root 
of  81  ^flS"  is  zfc  3  p'os. 

The   fifth   root   of   4 /a2    is  ^4   .   ]^a^=\^'^  .  ]^a^  :i= 

Art.  13S.  The  division  of  irrational  quantities  often  gives 
rise  to  fractions,  whose  numerators  and  denominators  are  both 
irrational.  In  such  a  case,  it  is  often  desirable  to  convert  the 
fraction  into  another  equivalent  to  it,  but  of  a   simpler  form. 

This  may  be  accomplished  by  multiplying  both  terms  of  the 
fraction,  by  any  quantity,  which  will  rtnder  oni   of  them  ra" 

ion  a] 


X.LV.  WITH    THE    RADICAL    SIGN.  205 

Thus,  if  both  terms  of  — _-  or  I  /     —  be  multiplied  by  v/«, 

we  have  — =,  or  if  both  be  multiplied  by  ^x,  we  have  V^^ 
\/ax  X 

.            V^ 
In  like  manner,  multiplying  both  terms  of  the  fraction  — 

b}  \/{a  +  xy\  gives        3  "  = ^"p^ = 

l/¥  .  \/(a-\-xY  __  ^b^(a-\-xY  ^   _^  ^~^     _ 

i ^=    i »  ^so,  —    1=  —     5— J     = 

a  +  x  a  +  x  J,  -i    —^ 


-       n- 


2"  X"  X 

n— 1 


X 

Since  the  product  of  the  sum  and  difference  of  two  quantities 
is  the  difference  of  their  second  powers,  Art.  33,  if  we  would 

render  the  denominator  of  — 1 _     rational,  we  multiply  both 

3-^2 

terms  of  the  fraction  by  3+i/2.      We  have  then  — ?- — -    =: 

V/2(3  +  v/2)  _      3v/2"H-2     _     3^/2+2 

(3~\/2)  (3  +  V/2)    ~    32_(^2")2  9^=2~     ~. 

3i/2+2        ,           ,  .      .                                                  a  —  \/'h 
-1 .    Also,  multiplymg  both  terms  of  the  fraction - 

7                                                _  a  +  t/j 

by  a  — 1/6,  we  have 1 . 


To  render  the  denominator   of. ^  _   rational 

^10—^2  —  ^3 

first  multiply  numerator  and  denominator  by  ^10-|-t/2  +l/3 
18 


206  IRRATIONAL    QUANTITIES  XLV 


V/30  +  »/6  +  3 
ivhich  gives  1 L__ ,  then  multiply  both  terms  of  thia 


last  bj  5-j-2i/6  ,  which  gives 


5—2/6 

5/30  +  2/180+11/6  +  2'; 


25  —  24 


=  5/30+12/5  +  11/6+27. 

Simplifications  of  this  kind  may  be  made  in  fractions  nvolvin^ 
Radicals  of  other  degrees  than  the  second ;  but,  except  when  the 
quantity  to  be  rendered  rational  is  a  monomial,  the  process  be- 
comes so  complicated  as  to  be  inconsistent  virith  the  design  oi 
this  treatise. 

Remark.  In  the  following  questions,  let  the  learner  simplify 
his  results,  when  it  can  be  done. 

1.  Add  /8    and  /50. 

2.  Add  /16^  and  /46. 

a  Add  /36  a2  y  and  /25y. 

4,  Add  /500  and  /108. 

5  Add  4/147  and  3/75. 

6  Add  3/1  and  2/^. 

7  Add  9/243"  and  10/363 

8  Add  12  ^i"  and  3^^. 

9.    Add  ^/a26"and  ^/46x4 
10.   Add  /12  +  2/27  and  3/75  —  9/48. 
11     Add  7^54  +  3^/^16  and  ^2  ^5^128. 
12L    Add  ^ST  —  2^24  and  /28  +2/63. 

13.  Add  /18a5  63  and  /50  a^  ¥. 

14.  Add  /45c3— /8073  and  /So^ 

15.  Add  y/    -p-  and-y/    S^ —\/    T^- 


XLV  WITH    THE    RADICAL    SIGH  2(0 

16.  From  y/50   subtract  ^8. 

17.  From  ^448"  subtract  ^112.' 
18  From  ^192  subtract  ^24. 

19.  From  5  ^20  subtract  3  y/45. 

20.  From  ^320  subtract  ^40. 

21.  From  ^f    subtract  \^^, 

22.  Fromy/8  subtract  2y/|". 

23.  From  ^72  subtract   3  y^^ 


24.  Fromy/SOa^a;  subtract  y/20  aS  x3. 

25.  From  S^o^y  subtract  2^a6T. 

26.  From  ^256  subtract  y^32. 

27.  From  ^^  subtract  \/\ 

28.  From  7^54  +  3  ^Te  subtract  5  ^128  —  ^%. 

29.  From  ^|  —\/l  subtract  y/l  —^\, 

30.  Multiply  3  ^2  by  2  ^2. 

31.  Multiply  ^2    by  ^8. 

32.  Multiply  ^2    by  ^4. 

33.  Multiply  ^o"  by  \/b, 

34.  Multiply  2 ^a6  by  3^ac. 

35.  Multiply  5 ^a^ c^    by  a ^ac, 

36.  Multiply  5^5  by  3y/a 

37.  Multiply  2  ^3  by  3^4! 

38.  Multiply  2 a  + 3^6  by  2 «  — 3^6, 
39  Multiply  7  +  3^12  by  3  +  4^2. 
40.  Multiply  3  +  ^5"  by  2  — ^5^ 


tub  IRRATIONAL    dUANTlTlEo.  XLV 

41.  Multiply  y/2  H-^3  by  2^2"  —y/f . 

3  &  — 

42.  Multiply  fci/a^  by  cy/fl. 

43.  Multiply  ^f  by  Yi". 

44.  Multiply  ly/f  by  3^/1: 

45    Multiply  together  ^2,  ^6    and  ^12. 

46.  Multiply  together  2^3^  3^4    and  6y/8. 

47.  Multiply  together  3 y/rt 6,  4^a6    and  2^m 

_  3   _  7  — 

48  Multiply  together  ^^^,  ^^7  and  ^6. 

49  Divide  6  ^a~  by  3  ^oT 

50  Divide  8  ^oT  by  2  ^a. 

51  Divide  S^x^  by  5^^^. 

52.  Diride  8^108  by  2^6. 

53.  Divide  ^^5"  by  ^^2. 

54.  Divide  ^7"  by  ^tT 

55.  Divide  3^^  by  2^^ 

56.  Divide  a  i/x  y  by  b  C/4:  c. 

57.  Divide  a +  y/6     by    a  —  ^h. 

Remark.  In  this  and  the  two  following  examples,  first  repre- 
lent  the  division,  and  then  simplify  by  rendering  the  denomina- 
ors  rational. 

58.  Divide  4 +  ^2"  by  3  +  ^27 

59.  Divide  ^3  by  3  +  ^3. 

3  _ 

GO    Find  the  2d  power  of  ya. 
61.    Find  the  2d  pow<ir  of  5  ^62". 

62  Find  the  2d  power  of  'i^'ab 

6  — 

63  Find  the  3d  power  of  4  ^a  z. 


XLVI.  NEGATIVE    EXPONENTS.  2119 

8  

64.  Find  the  4th  power  of  a i/b^  c^. 

65.  Find  the  5th  power  of  m^^c^. 

66.  Find  the  wth  power  of  z  y  »/a  b. 

67.  Find  the  mth  power  of  ^x  y. 
68  Find  the  3d  power  of  ^y/37 

69.  Find  the  4th  power  of  ^^6«. 

70.  FindtheSdpower  of  ^^24. 


71.  Find  the  5th  power  of  ^(a  +  b)\ 

72.  Find  the  2d  root  o{  ^^W^, 

73.  Find  the  3d  root  of  27 ^a6. 

74.  Find  the  3d  root  of  64 sj'am. 

75.  Find  the  4th  root  of  16  ^oH^^ 
"6.  Find  the  3d  root  of  \\/a^. 

77.  Find  the  5th  root  of  \/a  m. 

78.  Find  the  mth  root  of  ^(a  +  6)2. 


79,  Find  the  mth  root  of  \/\a  —  bf 

n  

80.  Find  the  mth  root  of  ^x  +  t/. 


81.   Find  the  3d  root  of  3^a  +  6. 


SECTION    XLV 


NEGATIVE    EXPONEITTS. 


Art.  133.  We  have  already  seen,  that,  to  divide  one  power 
of  a  quantity  by  another  power  of  the  same  quantity,  we  must 
subtract  the  exponent  of  the  divisor  from  that  of  the  dividend. 
Thus,  a'  divided  by  a^  gives  a^~^  =z  a^.      We  have  also  seen, 

18* 


5JiO  NEGATIVE    EXPONENTS  XLVl 


that  when  the  dividend  and  divisor  are  alike,  the  quot  ent  has 
zero  for  an  exponent,  and  is  equal  to  unity.     Thus,  —  .:=.  a^  z=z 

If,  however,  the  exponent  of  the  divisor  is  greater  than  that  of 
the  dividend,  the  quotient  will  have  a  negative  exponent.    Thus, 

In  order  to  understand  the  signification  of  negative  exponents, 

let  us  take  any  fraction  as  -g,  which  has  different  powers  of  the 

same  letter  for  its  two  terms,  but  in  which  the  exponent  of  the 
denominator  exceeds  by  1  that  of  the  numerator.    This  fraction, 

reduced  to  its  lowest  terms,  becomes  - ;  but  if  the  division  rep- 

a 

resented,  be  performed  by  subtracting  the  exponent  of  the  divi- 
sor from  that  of  the  dividend,  the  fraction  becomes  rt^~2  ir:  a~i. 

These  two  values  of  the  fraction  must  be  equal,  and  hence,  -  = 

a 

a 
Again,  take  the  fraction  — ,  in  which  the  exponent  of  the  de- 
nominator exceeds  that  of  the  numerator  by  2.     The  two  values 
of  this  fraction,  obtained    as   in   the   preceding   example,  are 

-r-  and  «~^ ;  therefore,  -^=.  a~^.     In  like  manner,  —  z=za^^\ 
—  —  a-4;  —  r=a~5.  and,  in  general, —-=  a~"*;  -; — t—ttz.  = 

Hencej  unity  divided  hy  any  quantity^  is  equal  to  the  same 
quantity  with  its  exponent  taken  negatively. 

Upon  the  principle  just  explained,  the  denominator  of  a  frac^ 
tion,  or  any  factor  of  the  denominator,  may  be  transferred  to  tht 
numerator,  care  being  taken  to  change  the  sign  of  the  exponent 
ffthe  quantity  thus  transferred. 


X.LVI  NEGATIVE    EXPONENTS.  2i  I 


Sam^           3  a  m2 

1         1           3am2                   _ 
6        c3              4 

''''    46c3    -       4       • 

4 

It  is  evident,  on  the  other  hand,  that  any  factor  of  the.  nwne- 
rator  having  a  negative  exponent^  7nai/ype  carried  to  the  df^aovii' 
natorf  if  the  sign  of  that  exponent  be  changed;  or,  token  any 
quantity,  integral  in  form,  contains  factors  having  negative  ex- 
ponents,  we  may  convert  them  into  a  denominator,  observing  merely 
to  change  the  signs  of  the  exponents. 

Thus, =:  ;  also,  a" 3  6-2 wi~^ 


4         ~4a2  65'         '  ~  cfib^m^' 

Art.  134r.  The  fundamental  operations  are  performed  on 
quantities  with  negative  exponents,  in  the  same  manner  as  if  the 
exponents  were  positive,  care  being  taken  with  regard  to  the 
rules  for  the  signs. 

Let  it  be  required  to  multiply  Sa^d  by  -^. 

By  the  usual  mode  of  multiplication,  the   result   would   be 

Sa^cd       Sa^c       ,.,.,,  o    o     j    i       « 

— - —  z=  — —-  ,  which   IS   the  same  as   3  a'^  c  d~^.     But    by 
d^  d 

transferring  d^  to  the  numerator,  and  then  multiplying   we  have 

3 a2 £? .  c  d-^  =:  3  a2  c  di-2  _  3 ^2 c  £^-1,  the  same  as  before. 

,     ,..  ^a^x      Sm^x^        3a2m-2x      Sa'^m^x^ 

In  like  manner,  - — -  .     „    ..     =  t . ^ 

4m2         Qa-^  4  ^ 

3.8a-*mx3       ^a~^mx^ 
~         479         ""  3        • 

46c 
Divide     ,    , —  by  3  m^  n  x^. 
m^n^x 

46c 
By  the  usual  method,  we  have  _    .    -   ,.     By  the  use  of  neg* 

46c 
ative  exponents,     - r-  3  m^n  x^  =i  A  b  c  m~^  n~^  x~^  -r 

3wi2nx*=  ^bcm~^n~^x~^. 


212  NEGATIVE    EXPONENTS.  XLVI. 

3a3  2;-2y-2  ^  9^4a;-3y-3 

7  ~  28         * 

The  third  power  of  ii~^  b'^   is   a~^b^;   the  fourth  root  of 

a~^  b^c^  is  a~^  b^  c^. 

Let  the  learner  perform  the  following  questions,  observing 
that,  in  the  ijiiultiplication  of  fractions,  all  the  factors  of  the  de- 
iiouimator,  except  such  as  are  numerical,  are  to  be  transferred 
to  the  numerator,  and  then  the  operation  may  be  performed  as 
usual ;  and  that,  in  dividing  by  a  fraction,  the  divisor  is  to  be 
inverted,  and  then  the  process  is  the  same  as  in  multiplication. 
Aflerwards  any  numerical  factors,  common  to  numerator  ard 
denominator,  are  to  be  suppressed. 
-1.  Multiply  m^n~^  by  7 m^  w ~ 2  a;2 
2.    Multiply  3a-46-2c3by  4a-26--3c2. 

a    Multiply  15 a^&^c  by  2a-*  6-T^c-i 

4.  Multiply  10^ abc  by  10-3  J  52  ^i 

5.  Multiply  3-1  a-2  6-3  by  3a2  63. 

6.  Multiply  ^^^^  by  362  c3w. 

2  a3  z  t/ 

7.  Multiply  7  m4r3  by  ^^^^5^. 

^     .,  ,  .  ,     7a6c2  ^     2lmx^y 

8  Multiply  -— ^-  by  ^    .  ,^  \. 

n      »T    ,  •    .      3(r7  +  6)2   ^     4(c  — 1/)2 

9  Multiply  -LJ^    by^A^,. 

10  Multiply  ^-'i^-  +  -ric-d)        ^SbHx+ynjnj-n) 
lu  Muitipjy  462(a:  +  y)  ^^  25(m  +  n)5^c  — rf)S 

1 1  Divide  6a^m^n  by  12  a^  m^  ». 

12  Divide  5  a^  z  -  ^  ^3  by  3  a2  x2  yS.  . 


3  a2  r3  «,4 

13.  Divide      ,^        by3q-3x-»mi 

46c        '' 

14.  Divide  4  a  6  c^  x^  by  o^sTa* 


XLVII.  INEQUALITIES  213 

,6  Divide '("  + ')^  ^^-^)  by '^"  +  ')'(^^^^' 

17  Divide  ^^^'(^  +  ^)'  by  ^^'  (^  +  ^)'  ^"~y) 

18.  Find  the  2d  power  of  3a-i  ft-^  c3. 

19  Find  the  3d  power  of  4  a -2  52^-3. 

20  Find  the  ^th  power  of  2  a^  6"  ^  c-f 

21.  Find  the  3d  power  of  10 m"*  z"  ^  yS 

22.  Find  the  2d  root  of  16a-2  6-^  cP. 

23.  Find  the  3d  root  of  8  or -9  63  c"  6. 

24.  Find  the  4th  root  of81a6-ic-2d-4. 

25.  Find  the  3d  root  of  27  a^  6"^  c-^. 

26.  Find  the  5th  root  of  3a-^  2y-^  7»- ^. 


SECTION    XLVII. 


[NEQUALITIES. 


A.rt.  13^.  Any  expression  which  indicates  that  one  of  two 
quantities  is  greater  than  the  other,  is  called  an  imquality. 
Thus,  a^by  which  is  read  a  greater  than  b,  and  m<^n,  which 
is  read  m  less  than  n,  are  inequalities. 

As  inequalities  frequently  occur  in  mathematics,  it  is  proper 
to  introduce  here  some  explanation  of  them. 

It  is  to  be  remarked,  that,  although  strictly  speaking  no  quan^ 
tity  can  be  less  than  zero,  yet,  in  the  theory  of  inequalities,  it  is 
convenient  to  consider  negative  quantities  less  than  zero,  posi- 
tive quantities  beinor  considered  greater  than  zero.  Moreover, 
a  negative  quantity  is  said  to  be  so  much  the  less,  in  proportion 
Ufc>  its  absolute  value  is  greater.    Thus,  0  ]>  —  2,  and  —  3  ^  —  7 


814  INEQUALITIES.  XLVIl 

With  a  few  exceptions,  the  principles  established  relative  to 
equations,  are  also  applicable  to  inequalities.  We  shall  proceed 
to  notice  these  principles  and  the  exceptions. 

The  quantities  separated  by  the  sign  ^,  are  called  memhers 
oftlic  inequality.  An  inequality  is  said  to  continue  in  the  same 
sense,  when  that  member  which  was  the  greater  previous  to  a 
particular  operation,  continues  so  afterwards;  and  two  inequali- 
ties are  said  to  exist  in  the  same  sense  with  regard  to  each  other ^ 
when  the  corresponding  members  are  the  greater  members. 
Thus,  a'^b  and  c^d  exist  in  the  same  sense,  because  the  first 
member  of  each  is  greater  than  its  second. 

X.  The  same  quantity  or  equal  quantities  may  he  added  to  both 
mchtbers,  or  subtracted  from  both  members  of  an  inequality ,  and 
the  inequality  will  continue  in  the  same  sense  as  before. 

Thus,  if  5^3,  by  adding  4,  we  have  5  +  4  ]>  3  -{-  4,  or 
9^7;    also,  if  a  >  6,  we  have  a  -\-  c^  b  -\-  c       Again,  if 

—  3>  — 7,  by  adding  8,  we  have  8— 3>8  — 7,  or  5>1; 
also,  if  —  a ^  —  6,  we  have  c  —  a^  c  —  b. 

Moreover,  the  inequalities  10]>7,  and  a^b,  give,  by  sub- 
traction, 10  —  5^7  —  5,  or  5 ]> 2,  and  a  —  c^b  —  c. 

Hence,  we  may  transpose  from  one  member  to  the  other  any 
term  of  an  inequality,  taking  care  to  change  its  sign ;  because 
that  is  equivalent  to  subtracting  the  same  quantity  from  both 
members,  or  adding  the  same  quantity  to  both  members.  Thus, 
?f  3  X  -|-  20  ^  40  —  x,  we  have,  by  transposition,  3  x  -|-  2;  ]>  40 

—  20,  or  4  x>  20. 

2.  The  corresponding  members  of  two  or  more  inequalities, 
existing  in  the  same  sense  with  respect  to  each  other,  may  he 
added,  and  the  resulting  inequality  will  exist  in  the  same  sense  as 
the  given  inequalities. 

Thufl,  by  adding  the  two  inequalities  5^3  and  15^7,  w<^ 
have  5  -f  15>3  +  7,  or20>10.  Also,  ifa>6,  c>f/,  and 
c  ^fy  ^ve  have  a  -\-  c  -\-  e'^  h  -\-  d  -\-f. 

3,  But  if  two  inequalities  existing  in  the  same  sense,  be  sub 
tr acted,  member  from  member,  the  resulting  inequality  will  not 
always'  exist  in  the  same  sense  as  the  given  inequalities. 


XL  VII.  INEQUALITIES.  215 

Indeed  the  resul  may,  according  to  circumstances,  be  an  ine- 
quality in  the  same  sense  as  those  given,  or  one  in  a  different 
Bense,  or  it  may  be  an  equation. 

Thus,  13  >  4  and  20  >  7  give,  by  subtraction,  20  —  13  >  7 
—  4,  or  7^3,  which  is  an  inequality  in  the  same  sense  as  the 
two  proposed. 

Again,  15  >  12  and  10  >  3  give,  by  subtraction,  15  —  10  <^ 
12  —  3,  or  5  <^  9,  an  inequality  in  the  opposite  sense  to  the  pro- 


Finally,  20>  17  and    12  >  9  give  20  —  12  =  17  —  9,  or 

8  =:  8,  an  equation. 

In  general,  let  a  ^  6  and  c'^  d;  then,  according  to  the  par 
ticular  values  of  a,  6,  c  and  rf,  we  may  have  a  —  c^  b  —  d, 
a  —  c<^h  —  rf,  or  a  —  c  =::  b  —  d. 

4.  The  two  members  of  an  inequality  may  be  multiplied  or  di 
vided  by  the  same  positive  quantity ^  or  by  equal  positive  quanti- 
ties, and  the  result  will  be  an  inequality  in  the  same  sense  as  the 
proposed. 

For  example,  multiplying  both  members  of  11^7  by  8,  we 
have  88 ^ 56.     Also,  if  a^b,  ac^b  c. 

Again,  dividing  both  members  of  35  ^  21    by  7,  we  have 
5^3.     Also,  if  am^cm,  we   have   a^  c;    and   if  m ^ w, 
m        n 
p       p' 

5.  Butf  if  both  members  of  an  inequality  be  multiplied  by  the 
same  negative  quantity,  or  by  equal  negative  quantities,  the  re- 
lult  will  be  an  inequality  in  a  sense  opposite  to  that  of  the  pro 
posed. 

Thus,  if  7^5,  and  we  multiply  by  — 3,  we  have  — 21  < 

—  15      Also,  if  a  ^  6,  multiplying  by  —  m,  we  have  —  am  << 

—  b  m.  In  these  examples,  the  sens«^  is  inverted,  because  a 
negative  quantity  is  less  in  proportion  as  its  absolute  value  is 
greater. 

6.  Htnce  it  follows,  that  the  sense  of  an  inequality  will  be  in- 


216  INEQUALITIES.  XLVIl 

verted,  if  all  the  signs  of  both  members  be  changed ;  because  thii 
is  the  same  as  multiplying  both  members  by  —  1. 

7.  Both  members  of  an  inequality,  if  they  are  positive  quanti^ 
ties,  can  be  raised  to  the  same  power,  and  the  result  will  be  an  in^ 
equality  in  the  same  sense  as  the  proposed. 

Thus,  from  7>2  we  have  72>22,  or49>4;  and  fJGto 
a^  b,  a'"^b'^. 

8.  But  if  both  members  of  an  inequality  are  not  positive,  and 
both  be  raised  to  the  same  power  denoted  by  a  whole  number,  the 
*'esulting  inequality  will  not  always  exist  in  the  same  sense  as  the 
proposed. 

Thus,  3  >  —  2  gives  3^  >  (—  2)2,  or  9  >  4,  in  the  same 
sense  as  the  proposed.  But  —  3  >  — ^5  gives  ( —  3)2  <;^  ( —  5)®, 
or  9  <^  25,  in  the  reverse  sense  of  the  proposed. 

9.  Roots  to  the  same  degree,  of  the  two  members  of  an  inequal- 
ity, may  be  extracted,  and  the  resulting  inequality  will  be  in  the 
same  sense  as  the  proposed. 

Thus,   27  >  8   gives   ^27  >  ^8^  or,  3  >  2,  and,  in   gen 

n  ,  n 

eral,  a^b  gives  i/a  ^  r/6. 

Tf  the  root  be  of  an  even  degree,  it  is  necessary  that  both 
members  of  the  given  inequality  be  positive ;  otherwise  one  or  both 
of  the  roots  would  be  imaginary,  and  they  could  not  be  com- 
pared. 

Art.  136.  There  are  some  problems,  the  solution  of  which 
involves  the  principles  of  inequalities.  The  following  are  of  this 
kind. 

1.  Three  times  a  certain  number  added  to  16,  exceeds  twice 
that  number  added  to  24,  and  two  fifths  of  the  number  added  to 
5  is  less  than  11.     Required  the  number. 

Let  X  represent  the  number  ;  then  Sx  -\-  16^2x-|- 24,  and 

2  X 

\-  ^  <^  11.     The  first  inequality,  by  transposition  and  re- 

5 

duction,  gives  z  ^  8.     The  second,  multiplied  by  5,  becomes 


XLVIII.  EQUIDIFFERENCE.  217 

2z  -\-25<^^o,  which,  by  transposition,  reduction,  and  division, 
gives  x<^  15.  Any  number,  therefore,  entire  or  fractional,  which 
is  greater  than  8«and  less  than  15,  will  fulfil  the  conditions  of  the 
question. 

2.  Says  A  to  B,  I  have  an  exact  number  of  dollars  in  my 
|j!jrs«;  if  I  had  twice  as  many  and  $10, 1  should  have  more  than 
$19;  but  if  I  had  three  times  as  many,  my  number  would  be  less 
than  the  number  I  now  have  increased  by  $41.  Required  the 
number  of  dollars  in  his  purse. 

3.  A  certain  number  divided  by  17  gives  an  entire  quotient, 
which  quotient  increased  by  2,  exceeds  4 ;  but  if  the  number  be 
multiplied  by  2,  and  the  product  be  increased  by  4,  the  result 
will  be  less  than  the  number  itself  increased  by  56.  What  la 
the  number? 


SECTION    XLVIII. 


EQUIDIFFERENCE. 


Art.  137.  The  difference  between  two  quantities  is  some 
times  called  their  arithmetical  ratio,  or  ratio  oy  subtraction 
Thus,  the  arithmetical  ratio  of  9  to  7  is  9  —  7  or  2,  and  that  of 
a  to  6  is  a  —  6. 

Four  quantities,  such  that  the  difference  between  the  first  and 
second,  is  the  same  as  that  between  the  third  and  fourth,  consti- 
tute what  is  called  an  equidifference,  sometimes  called  also  an 
arithmetical  proportion.  Thus,  9,  7,  5  and  3  form  an  equidiffer- 
ence; for  9  —  7  =  5  —  3.  This  is  sometimes  expressed  thu 
9.7:5.3,  in  which  one  point  denotes  difference,  and  two  points 
denote  equality.  But  this  notation  is  objectionable,  because 
these  characters  are  sometimes  used,  the  one  to  represent  multi 
plication,  and  the  other  division. 

In  like  manner,  if  the  quantities,  a,  6,  c  and  c?,  are  such  thai 
a  —  bzrzc  —  rf,    or    —  a-\-bz=:  —  c -}- d,  these  four  quaniiiiei 
constitute  an  equidifference. 
19 


218  EQVIDIFFERENCE.  XLVIU 

The  quantities,  «,  b,  c,  d,  are  called  terms  of  the  equidiffe^ 
euce.  Also,  a  and  d,  the  first  and  last  terms,  are  called  the  ex^ 
tremeSf  because  they  occupy  the  extremities ;  b  and  c,  the  second 
and  third  terms,  are  called  the  means,  because  they  occupy  the 
middle  in  the  equidifference. 

Remark.  In  the  definition  of  equidifference,  it  is  supposed, 
that,  if  the  second  term  is  greater  than  the  first,  the  fourth  is 
greater  than  the  third ;  but,  if  the  second  is  less  than  the  first 
the  fourth  is  less  than  the  third. 

From  any  equidifference,  a  - —  b  =:  c  —  d,  or,  —ra-{-b=z 
—  c  -\-d,  we  deduce,  by  transposition,  a-\-  d  =i  b  -\-  c;  also, 
«  =r  6  -f-  c  —  d,  and  d=ib  -\-c  —  a,  b=:  a-\-d  —  c,  and  c  =r. 
a-\-d — b.     Hence, 

In  any  equidifference,  the  sum  of  the  means  is  equal  to  the  sum 
of  the  extremes.  Moreover,  either  mean  is  equal  to  the  sum  of  the 
extremes,  diminished  by  the  other  mean  ;  and  either  extreme  is 
equal  to  the  sum  of  the  means,  diminished  by  the  other  extreme. 

Suppose  we  have  a-\-  dz=b-\-c;  by  transposition  we  obtain 
a  —  bz=:  c  —  d. 

Therefore,  if  the  sum  of  two  quantities  is  equal  to  the  sum  of 
two  other  quantities,  the  first  two  may  be  made  the  means,  and  the 
last  two  the  extremes,  or  the  reverse,  of  an  equidifference. 

When  three  quantities,  a,  b,  c,  either  increasing  or  decreasing, 
are  such  that  the  difference  between  the  first  and  second  s  equal 
to  that  between  the  second  and  third,  that  is,  a  —  bz=.b  —  c, 
they  constitute  what  is  called  a  continued  equidifference,  and  the 
quantity  b  is  called  the  arithmetical  mean  between  a  and  c.  Thus, 
3,  5,  and  7,  or  12,  8,  and  4  form  a  continued  equidifference. 

Take,  for  example,  a  —  6  =  6  —  c.      From  this  we  deduce 

(I    I    c 
i  =  — ^—  ;  alsOj  az=z^b  —  c,  and  cz=z^b  —  a.     Hence, 

In  any  continued  equidifference,  the  mean  is  half  the  sum  of  the 
extremes,  and  either  extreme  is  found  by  subtracting  the  other  cav. 
treviefrom  twice  the  mean 


XLIX  PROPORTIONS.  21P 

1    The  means  of  an  equidifference  are  10  and  12,  and    he 
known  extreme  is  6.     Required  the  other  extreme. 

2.  The  extremes  of  an  equidifference  are  7  and  4^,  and  one 
of  the  means  is  6.     What  is  the  other  mean  ? 

3.  The  means  of  an  equidifference  are  8  and  12,  and  t\e  last 
term  exceeds  twice  the  first  by  5.     Required  the  extremes. 

4.  In  a  continued  equidifference,  the  extremes  are  10  and  15^. 
Required  the  mean. 

5.  In  a  continued  equidifference,  the  mean  is  7  and  onf  ex- 
tieme  is  8.     Required  the  other  extreme. 

6.  The  mean  of  a  continued  equidifference  is  14,  and  the  third 
term  exceeds  the  first  by  8.     Required  the  extremes. 


SECTION    XLIX. 


RATIO    AND    PROPORTION. 


Art.  138.  The  quotient  arising  from  the  division  of  one  quan- 
tity  by  another,  whether  the  division  can  be  exactly  performed 
or  can  only  be  expressed,  is  called  the  ratio  of  these  quantities. 
It  is  sometimes  called  ratio  hy  division y  or  geometrical  ratio. 
But  when  the  word  ratio  simply  is  used,  it  signifies  ratio  result- 
ing from  division. 

A  ratio  is  most  appropriately  expressed  in  the  form  of  a  frac- 
tion.    ThuiS,  f  is  the  ratio  of  3  to  5,  and  j-  is  that  of  a  to  b. 

An  equation  formed  by  two  equal  ratios,  is  called  2i  proportion. 
Sometimes  the  term  geometrical  proportion  is  used,  to  distin- 
guish it  from  arithmetical  proportion  or  equidifference.     Thus, 

3        9         J  a        c 

-  zr:  --,  and  7  =  -  are  proportions. 

7        21  0        a 

For  the  sake  of  convenience  in  writing  and  printing,  most  au« 
thors  express  division  by  the  sign  :  ,  placed  between  the  quanti- 
ties, and,  instead  of  the  sign  =,  use  the  sign  :  :       Thus,  a:b: . 


220  PROPORTIONS  XLIX 

c:d  18  read,  a  is  to  i  as  c  is  to  d^  and  is  the  same  as  y  =^  ~i 

o  a 

The  signification  in  both  cases,  is,  that  a  divided  by  b,  is  equal 

to  c  divided  by  d.     In  this  treatise  points  will  sometimes  be  usea 

to  denote  division,  but  the  sign  =:  will  always  be  preferred  rather 

than  : :  . 

In  any  proportion,  a  :  b  z=z  c  :  d,  the  quantities  a,  b,  c,  and  </, 
are  calle^l  the  terms  of  the  proportion.  The  two  quantities  a 
and  b  are  called  the  terms  of  the  first  ratio,  c  and  d  those  of  the 
second. 

Moreover,  a  and  c  are  called  the  antecedents  of  the  propor- 
tion, a  being  the  antecedent  of  the  first  ratio,  and  c  that  of  the 
second ;  6  and  d  are  called  the  consequents  of  the  proportion,  b 
being  the  consequent  of  the  first  ratio,  and  d  that  of  the  second. 
Also,  a  and  d  are  called  the  extremeSy  b  and  c  the  means  of  the 
proportion. 

These  names  are  derived  from  the  position  in  which  the  terms 
stand  with  respect  to  each  other,  when  the  division  is  indicated 
by  points.  Antecedent  signifies  going  before,  and  consequent, 
following  after.  Thus,  in  the  ratio  a.b,  a  precedes  and  b  fol- 
lows after  it.  The  signification  of  the  words  means  and  extremes 
has  already  been  explained. 

Art.  139.  There  are  several  important  properties  of  propor- 
tions, which  we  shall  now  proceed  to  demonstrate. 

1.  Take  any  proportion,  a  •  6  =  c  :  rf,  or  -  =  -^  If  we  mul- 
tiply the  proportion  in  its  second  form,  by  the  denominators  b 
and  c?,  we  have  adz=bc.  But  a  and  d  are  the  extremes,  and  b 
and  c  the  means.     Hence, 

In  any  proportion  the  product  of  the  means  is  equal  to  the  pro 
duct  of  the  extremes. 

2.  Suppose  we  have  ad=zbc.     Dividing  both  members  by  i 

a        c 
iind  d.  we  have  Y=^,ora:6  =  c:rf.     Hence, 
0       a 


KLTX  PROPORTIONS,  2*i* 

If  the  product  of  two  quantities  is  equal  to  the  product  oj  twa 
otner  quantities^  the  two  factors  of  either  product  may  be  made 
the  means  ^  and  the  two  factors  of  the  other  product  ^  the  extremes 
of  a  proportion. 

3.  Any  three  terms  of  a  proportion  being  given,  we  can  always 
find  the  reinaining  one.     For,  take  any  proportion,  a :  6  =  c  :  cf, 

or  7-  =  -  ,  which  gives  ad=zbc;  hence,  by  division,  a=z  -j, 
b        d  °  d 

f/rr:  — ,  b  i=  — ,  and  c  =  -r-.     Therefore, 
a  c  b 

In  any  proportion,  either  mean  is  equal  to  the  product  of  tlie 
extremes^  divided  by  the  other  mean ;  and  either  extreme  li  equal 
to  the  product  of  the  means ,  divided  by  the  other  extreme. 

From  this  it  follows,  that, 

If  three  terms  of  one  proportion  are  respectively  equal  to  the 
three  corresponding  terms  of  another  proportion,  the  remaining 
term  of  one  must  be  equal  to  the  remaining  term  of  the  other. 

4.  The  proportion,  a.b^zb  :  c,  in  which  the  two  mean  terms 
are  the  same,  is  called  a  continued  proportion,  and  6  is  called  a 
mean  proportional  between  a  and  c.    This  proportion  gives  b^  = 

a  c,  and  b  =  r/a  c.     Hence, 

The  mean  proportional  between  two  quantities,  is  equal  to  the 
second  root  of  their  product. 

From  this  it  follows,  that. 

If  the  second  power  of  any  quantity  is  equal  to  the  product  of 
two  others,  the  first  quantity  is  a  mean  proportional  between  thm 
last  two. 

For  the  equation,  a  c  =  b^,  gives  a:bz=b  :c. 

5.  Let  there  be  given  a:b:=c:d.     (1) 
This  produces  ad=:  be.     (A) 

Dividing  both  members  of  equation  (A)  by  c  and  d,  we  hav* 

—  =  —,  or  a  :  c  =  b  :  d.     (2) 
c         d  ^ 

19* 


222  PROPORTIONS.  XLIX 

Dividing  equation  (A)  by  a  and  6, 

d        c 

—  =1  -y  or  d :  b  =  c  :  a.     (3) 

Changing  the  order  of  the  ratios  in  proportion  (1), 

c:d=a:b.     (4) 
Changing  equation  (A)  member  for  member,  and  then  di?» 
ding  by  a  and  c, 

-  =  -  ,  or  b  :  a=id:  c.     (5) 
a        c 

Comparing  proportions  (2),  (3),  (4),  and  (5),  with  the  given 
proportion  (1),  we  infer,  that, 

Iji  any  proportion  the  means  may  exchange  places ;  the  extremes 
may  exchange  places ;  the  extremes  may  be  made  the  means^  and 
the  means  the  extremes ;  both  ratios  may^  at  the  same  time,  be  in- 
verted, that  iSy  the  antecedent  and  consequent  of  each  ratio  may 
exchange  places. 

Indeed,  in  a  given  proportion,  any  change  may  be  made  in  the 
order  of  the  terms,  provided  that,  in  each  arrangement,  the  pro- 
duct of  the  means  being  put  equal  to  the  product  of  the  extremes, 
the  same  equation  is  produced,  as  that  arising  from  the  given 
proportion.  The  same  proportion,  therefore,  admits  of  eight 
forms,  viz  •     . 

a:bz=c:d;  a:  c  =  b  :d; 
d:b=zc:a;  d:c=ib:  a; 
b  :a=zd:  c;  b  :dz=a:c; 
c  :  a=.d :  b ;  c  :  d=z  a  :b; 
for  each  proportion  gives  ad=:zbc. 

6.  Since  the  value  of  a  fraction  is  not  changed,  when  both  nu- 
merator and  denominator  are  either  multiplied  or  divided  by  the 
same  quantity,  it  follows,  that. 

In  a  proportion,  we  may  multiply  or  divide  both  terms  of  either 
ratio  by  the  same  quantity,  and  we  may  multiply  or  divide  all  the 
terms  of  a  proportion  by  the  same  quantity,  without  disturbing  the 
proportion.      We  may  also  multiply  or  divide  both  terms  of  th§ 


XLIX.  PROPORTIONS.  223 

-first  ratio  hy  one  quantity ^  and  both  terms  of  tht  second  ratio  by 
another  quantity ^  or  we  may  multiply  both  terms  of  one  ratio  by 
any  quantity j  and  divide  both  terms  of  the  other  ratio  by  the  same 
or  a  different  quantity,  without  disturbing  the  proportion. 

a         c 
Thus,  if  a  :  6  =  c  :  c?,  or  —  =r  —  ,  we  have 
o        a 

am  :bmz=.  c:d;  a:  b  =  cn  :  dn; 

a       b  J  A        «     ^  *  • 

mm  n     n 

.  .abed 

am\  bm-=.  cm'.  dm\  —  :  —  r=:  —  :  — . 
m      7n        m       m 

Ai  t  .abed 

Also,  a  m  :  0  m  =1  c  n  :  an ;  — :  —  =  -  :-; 
mm       n      n 

,  c       d      o       b  , 

am:  b.n  =z:  —  :  —  ;   —  :  —  =:  en:  an. 
n       n      m      m 

7.  Both  of  the  antecedents  or  both  of  the  consequents  of  a  nrth 
portion,  may  cither  be  multiplied  or  divided  by  the  same  quauiity 
or  by  equal  quantities,  without  disturbing  the  proportion. 

Thus,  \{  a:b=:c:d,  ox  -  =1  — ,  we  have 
0         a 

am  ■    cm  ,  . 

—r-z=.—z-,oYam:b=zcm:d\  , 

b  d 

- —  =  rr-,  OT  a:bn  =  c:  dn. 
bn       dn 

Ai         ^      A         ^     ^  *  ^ 

Also,  —  :o=  -:a;  a:  -  =z  c  :  -. 

m  m  n  n 

The  reason  is  obvious,  for  these  several  results  are  produced 
by  multiplying  or  dividing  equal  fractions  by  the  same  quantity 

8.  Suppose  we  have  the  two  proportions, 

a  :  b  z=i  c  :  d,  and  a:b  =im:n. 
Then,  according  to  ax.  7,  we  have 
c  :  d=m:n.     Hence, 
If  two  proportions  have  a  common  ratio,  or  a  ratio  in  one  pro^ 
portion  equal  to  a  ratio  in  the  other,  the  two  remaining  ratios  are 
tqualf  and  may  form  a  proportion. 


224  PROPORTIONS.  XLIX 

9    Suppose  we  have  the  two  proportions 

a  :  b  z=:  c  :  dj  and  a  :  m  r=r  c  :  w,  in  which  the  aiStecedents 
are  alike.     By  changing  the  means  in  each,  we  have 

a:  c  =  b  :  dj  and  a:  c:=m:n;  consequently,  on  account 
of  the  common  ratio,  a  :  c,  we  have 

b:dz=m:n;  hence,  b:m=id:n.     Therefore, 
If  in  f^vo  proportions,  the  antecedents  are  alike  or  equal,  thi 
ionsequents  will  form  a  proportion. 
Suppose  now  that  we  have 

a  :  b  =z  c  :  d,  and  m  :  b  =z  n  :  d,  two  proportions  in  which 
the  consequents  are  alike..  Changing  the  meaus  in  each,  we 
have 

a:  c=::b  .d,  and  m:n=zb  :  d.  Consequently,  on  ac- 
count of  the  common  ratio, 

a  :  c=zm:n;  hence,  a  :m=zc:n.     Therefore, 
tfin  ttoo  proportions  the  consequents  are  alike  gr  equal,  the 
fintecedents  will  form  a  proportion. 

a        c 

10.  Suppose  a  :  b  z=i  c  :  d,  or  -  =:  — . 

0        a 

Adding  to  or  subtracting  from  both  members  of  the  equation 

any  quantity  m,  and  reducing  to  a  common  denominator,  we 

have 

a^hbrn       Czhdm  ,    .         ,  ,     .        _, 

7 rr — ,  or  a:izbm:b=zcdzdm:d. 

o  a 

The  last  proportion  becomes,  by  changing  the  means, 
a^hbrn:  czhdm=^b  :d=:a:  c   (1); 
since,  from  the  given  proportion,  these  last  two  ratios  are  equal 

If,  in  the  given  proportion,  a  :  b  =z  c  :  d,  both  ratios  be  in- 
verted, the  proportion  becomes 

b  :  a  =:  a  :  c,  or  -  =  -. 
a        c 

AVlding  to  or  subtracting  from  both  members  of  this  €quation 

any  quantity  wz,  and  reducing  to  a  common  denominator,  we 

have 


XLIX.  PROPORTIONS.  225 

h^am       d^cm  ,    .  ,  ,  ... 

— = ,  or  orh  awi .  a  =  ai  crw  :  6    which 

a  c 

if  the  means  be  changed,  becomes 

b:^am  :  d:±zcmz=  a:  c=zb  :  d.     (2) 

Comparing  proportions  (1)  and  (2)  with  the  given  proportion, 
we  infer,  that. 

In  any  proportion^  thejirst  antecedent  plus  or  minus  any  nuTn- 
her  of  times  its  consequent,  is  to  the  second  antecedent  plus  or 
minus  the  same  number  of  times  its  consequent,  also  the  first  con- 
sequent  plus  or  minus  any  number  of  times  its  antecedent,  is  to 
the  second  consequent  plus  or  minus  the  same  number  of  times  its 
antecedent,  as  tJie  first  term  is  to  the  third,  or  as  the  second  is  to 
the  fourth. 

11.  From  the  proportion  b^am:  rfrb cm  =  a:  c,  which  was 
obtained  above,  we  have,  by  taking  the  plus  sign, 

b-{-am:d-{-cm=za:  c;  by  taking  the  minus  sign, 
6  —  am:d  —  cm  =  a:  c    hence, 

b-{-am:d-\-cmz=b—   amid — cm;  making  »i=:l, 
b-\-a:d-\-c  =  b  —  a:  J — c ;  changing  the  means, 
b-\-a:  b  —  a=.d-\-c  :  d  —  c. 

From  the  last  two  proportions,  we  infer,  that, 

In  any  proportion,  the  sum  of  the  first  two  terms  is  to  the  sum 
of  the  last  two,  as  the  difference  of  the  first  two  terms  is  to  the 
difference  of  the  last  two ;  also,  the  sum  of  the  first  two  terms  is 
to  their  difference,  as  the  sum  of  the  last  two  terms  is  to  their  dif- 
ference. 

Remark.  It  is  manifest  that  the  last  two  proportions  might 
be  written  thus : 

a-\-b:c-\-d=:a  —  b  :  c  —  d,  and 

a-\-b:  a  —  bz=ic-^d:  c  —  d;  for  these  are  decuced  from 
those  proportions,  in  one  case,  by  changing  the  signs  of  both 
numerator  and 'denominator  of  a  fraction,  and,  in  the  other,  bj 
c  langing  the  signs  of  both  members  of  an  equation. 

12.  Let  the  proportion,  a  :  b  :=  c  :  d,  be  given. 


236  PROPORTIONS.  XLIX 

By  changing  the  means,  we  have 

a  :  c  z=  b  :  df  or  -  =  J  ;  whence, 
c         a 

a:±zcm       b^dm  ,  i    .     »         « 

z=  ; — ,  or  azhcm:  cz=z  b-±zdm  :o. 

c  d 

(  hanging  the  means  in  the  last  proportion, 

a^cm\b^dm:=i  c\dz=.a'.b,     (1) 

By  making  the  means  the  extremes,  and  the  extremes  the 
means,  in  the  given  proportion,  we  have 

c'.a:=:d:b,  or  —  =  — ;  whence, 
a         b 

czhani:  a  =  dzkzbm  :b;  changing  the  means, 

czham:  dzhbm:=a  :b=:c  :  d.     (2) 

Comparing  proportions  (1)  and  (2)  with  the  given  proportion 
we  infer,  that. 

In  any  proportion  ^  the  first  antecedent  plus  or  minus  any  num- 
ber of  times  the  second^  is  to  the  first  consequent  plus  or  minus  the 
same  number  of  times  the  second,  also  the  second  antecedent  plus 
or  minus  any  number  of  times  the  first,  is  to  the  second  consequent 
plus  or  minus  the  same  number  of  times  the  first,  as  either  ante* 
cedent  is  to  its  consequent. 

13.  By  making  m=l,  in  proportion  (2)  of  number  12,  we 
have 

czha:d::hb=za:b:=zc:d    (I);  taking  the  sign  -|-, 

c-{-a.d-\-b=:a:b;  taking  the  sign  — , 

c  —  aid  —  b=za:b;  whence, 

c-^a:d-{-b=:c  —  aid — b  (2);  changing  the  means, 

c-\-aic  —  a  =  d-\-bid — b  (3). 

The  last  two  proportions  may  evidently  become 
a-\-cib  -\-d=z  a  —  cib  —  d,  and 
a-{-cia — c=:b-\-dib — d. 

Comparing  proportions  (1),  (2),  and  (3)  with  the  given  proi 
portion,  aib^rcid.  we  infer,  tha*, 


XLIX.  PROPORTIONS.  221 

In  any,  proportion,  the  sum  or  difference  of  the  antecedents,  is 
to  the  sum  or  difference  of  the  consequents,  as  either  antecedent 
is  to  its  consequent ;  the  sum  of  the  antecedents  is  to  the  sum  of 
the  consequents,  as  the  difference  of  the  antecedents  is  to  the  dif- 
ference of  the  consequents ;  also,  the  sum  of  the  antecedents  is  to 
their  difference,  as  the  sum  of  the  consequents  u  to  their  differ' 
(Yice. 

14.  If  in  any  proportion,  the  antecedents  are  alike  c^  equat, 
the  consequents  must  be  equal;  also,  if  the  consequents  are  alike 
or  equal,  the  antecedents  must  he  equal. 

For  equal  fractions  having  equal  numerators,  must  have  equal 
denominators;  and  equal  fractions  having  equal  denominators, 
must  have  equal  numerators. 

Thus,  if  a  :  b  =z  a  :  m,  then  bz=:m;  or  if  a  :  6  =  c  :  m,  and 
a  =z  c,  then  h  z=.  m.  Also,  a  :  m  z=.  c  '.  m  gives  a  :=.  c,  and 
a'.bz=.c:m  gives,  upon  the  supposition  that  b=:m,  az=zc. 

It  is  moreover  evident,  that, 

If  the  second  term  is  greater  than  the  first,  the  fourth  must  be 
greater  than  the  third,  and  conversely ;  and  if  the  first  two  terms 
are  equal,  the  last  two  must  also  be  equal. 

15.  Suppose  we  have  a  series  of  equal  ratios,  viz . 

Let  q  represent  the  value  of  each  of  these  fractions ;  th^n, 

I  =9y-^~9^J  =  9y^  =  9-     By  multiplication, 

a=.bq,  czzzdq,  e  ^=fq,  g=^hq.     Adding  these  equations 
a^c  +  e-{-g  =  bq  +  dq+fq  +  hq,  or 
a^c  +  e-\-g=^(b-}-d-]-f-\-h\q    Dividing  by  6  +  c?+/+ A 
a-\-c-\-e4-g  a         ^ 

a-\-c-{-e-\-g:b-\-d  +/+  hz=  a:b  =  c:dz=e  :/=  g :  h. 
Hence, 


228  PROPORTIONS.  XlilX 

In  any  series  of  equal  ratioSy  the  sum  of  the  antecedents  is  ta 
the  sum  of  the  consequents ,  as  any  one  of  the  antecedents  i>  to  its 
consequent. 

a  c 

16.  J{a  :  b  =:  c  :  d,  and  e  :f=  g  :  h,  that  is,  -  =  — ,    and 

-^  =  ^,  by  multiplying  these  two  equations  together,  we  obtain 

lJ^Jf,^^^<^^''^f=^S''dh. 

The  same  result  would  have  been  obtained  by  multiplying  to- 
gether the  corresponding  terms  of  the  two  given  proportions. 
This  is  called  multiplying  the  proportions  in  order.     Hence, 

If  two  or  more  proportions  are  multiplied  in  order ,  the  result 
will  form  a  proportion. 

From  this  it  follows,  that. 

If  proportions  are  divided  in  order,  the  result  will  form  a  prO' 
portion. 

a         c 

17.  Given,  a  :  b  =  c  :  d^  or  -  =  —. 

0         a 

Raising  both  members  to  any  oower  denoted  by  rw,  we  have 
or       c"* 

Therefore, 

Similar  powers  of  proportional  quantities  form  a  proportion. 
From  this  it  follows,  that. 

Similar  roots  of  proportional  quantities  will  form  a  preptn 
tiom. 


PROGRESSION    BY    DIFFEREI^CE.  2^9 


SECTION    L. 

PRyGRJCSSION    BT    DIFFBREHTCB. 

Art.  140.  A  series  of  quantities,  such  that  eaci  is  greater 
ihan  that  which  immediately  precedes  it,  or  such  that  each  ex- 
ceeds that  which  immediately  follows  it,  by  the  same  quantity, 
"ionstitutes  what  is  called  a  progression  hy  differencCf  or  aritltr 
nrtical  progression. 

Thus,  the  natural  numbers,  1,  2,  3,  4,  &c.,  form  such  a  pro- 
gression, since  the  difference  between  any  two  contiguous  num- 
)e.rs  is  unity. 

Progression  by  difference  may  be  either  increasing  or  decreas- 
<ng.  The  series  3,  5,  7,  9,  &/C.  is  an  increasing,  but  20, 18,  16, 
(4,  &.C.  is  a  decreasing  progression. 

To  exhibit  a  progression  generally,  let  a  be  the  first  term,  and 
'  the  common  difference ;  then,  if  the  progression  be  increasing, 

t  3d  3d  4th  Sth 

(,  {a  -\-  d),  (a  +  2  d)y  (a  +  3  d),  (a  +  4  d),  &c.,  will  be  succes- 
ave  terms  at  the  commencement  of  the  series;  or,  if  the  progres- 

Irt  2d  3d  4th  Sth 

6ion  be  decreasing,  a,  {a  —  d),  {a  —  2  c?),  (a  —  3  c?),  (a  —  4  c?), 
dt/C,  will  be  the  successive  terms. 

By  examining  these  series,  we  perceive,  that  if  the  progression 

s  increasing,  the  second  term  is  found  by  adding  once  the  com- 
mon difference  to  the  first  term ;  the  third  term  is  found  by  add- 
ing twice  the  common  difference  to  the  first  term ;  the  fourth,  by 
adaing  three  times,  and  the  fifth,  by  adding  four  times  the  com- 
moi»  difference,  to  the  first  term.  But,  if  the  progression  is  de- 
ceasing, we  subtract  from  the  first  term,  once,  twice,  three  times, 
fuaf  times  the  common  difference,  to  find  the  second,  third, 
/uiifth,  fifth  terms.  That  is,  in  all  cases,  the  difference  is  multi 
Dij'jd  by  a  number  less  by  1  than  that  which  marks  the  place  of 
the  rerm,  and  the  product  is  either  added  to,  or  subtracted  from, 

h«i  ^rst  term.    Hence,  if  in  addition  to  the  notation  used  above, 
20 


230  PROGRESSION    BY    DIFFERENCE.  I, 

n  denote  the  number  of  terms,  and  /  the  last  terra,  we  shall  hav« 
the  formula 

l=za-^-{n  —  l)di  in  an  increasing  progression ;  and 
l=.a  —  (n  —  l)d,  in  a  decreasing  progression. 

If  we  use  the  double  sign  ri=,  the  general  formula  for  the  last 
term  is, 

I=a^{n—l)d.     Therefore, 

To  Jind  the  last  term,  multiply  the  common  difference  by  thi 
number  of  terms  minus  one,  and  add  the  product  to  theffrst  term 
if  the  progression  is  increasing ^  or  subtract  the  product  from  the 
first  term  if  the  progression  is  decreasing. 

Ex.  1.    Find  the  10th  term  of  the  progression,  3,  7,  11,  &c. 

In  this  example,  a  =  3,  c?z=4,  and  nzrzlO;  by  substituting 
these  values  in  the  formula,  l=ia-\-(n —  \)d,  we  have  /r=  3  + 
(10— 1)4  =  3  +  9.4  =  3  +  36  =  39.  The  last  term  therefore 
is  39. 

Ex.  2.    Find  the  8th  term  of  the  series,  50,  48,  46,  &c. 

We  have,  in  this  case,  Z=50  — (8  — 1)2  =  50  — 7  .  2  = 
50  —  14  =  36. 

Art.  14:1.  We  wish  now  to  find  a  formula  for  the  sum  of  any 
number  of  terms  in  progression  by  difference.  For  this  pur- 
pose, let  S  denote  the  sum  of  all  the  terms  in  the  progression,  a, 
a-\-d,  a  +  2  (?,  &c.     Then, 

Itt  2d  3d  4th  nth 

S=a+(a  +  d)  +  (a+2d)  +  {a  +  3d)+ +1     (1) 

If  we  begin  with  the  last  term,  it  is  evident  that  the  successive 
terms  of  the  same  progression  will  be  Z,  / — d,  I — 2  c?,  / — 3rf, 
&c.     Hence, 

nth         (n— l)th  (n_2)th  (n_3)th  .it 

5f=/+(/— ^)  +  (/— 2^  +  (/— 3rf)+ +a.    (2) 

Remark.     The   erms  cannot  all  be  written,  unless  some  defi- 
nite va.ue  be  given  to  n.     Points  are  therefore  used,  to  suppl 
.he  place  of  the  indefinite  number  of  terms  which  are  omitted 

Adding  the  equations  (1)  and  (2),  we  have 


L.  PROGRESSION    BY    DIFFERENCE.  231 

9S=(a  +  l)  +  {a  +  l)  +  (a  +  l)  +  {a  +  l)  +  ...  +  {a  +  l)i 
or,  since  the  quantities  included  in  the  several  parentheses  aie 
the  same,  and  since  n  represents  the  number  of  terms, 

2S=n(a-\-l);  and,  therefore, 

S="±±'X     Hence, 

The  sum  of  any  number  of  terms  in  progression  hy  difference^ 
is  found,  by  multiplying  the  sum  of  the  first  and  last  terms  by 
half  the  number  of  terms,  or  by  multiplying  half  the  sum  of  the 
first  and  last  terms  by  the  number  of  terms. 

By  substituting  the  value  of  /  in  the  formula  just  found,  we 

can  obtain  another  formula  for  S.    For,  since  l=za^(n —  \)d, 

o.      n[a-|-adc(» — 1)^1        2nadbw(n  —  i)d 
we  have  8=z  — ^ — ' ^ '—^  =  ^ '—  ,    or, 

nd(n — I) 
S=na±i ^— \ 

In  the  first  formula  for  the  sum,  it  is  necessary  to  find  I  before 
we  can  findiS^;  but,  by  the  second  formula,  <S^can  be  found,  when 
a,  d  and  n  are  known,  without  previously  finding  /. 

Ex.     Find  the  sum  of  12  terms  of  the  series,  7,  9,  11,  &c. 

In  this  example,  a  =:  7,  c?  =  2,  and  n  z=.  12.  We  first  find 
the  last  term  and  then  the  sum.  By  substituting  in  the  formula  for 
Z,  we  have  /=7  +  (12  —  l)2i=29.    Then  substituting  in  the  first 

formula  for  S,  we  have  8=  ^^(^  +  ^^)  —  6  .  36  =  216. 

By  using  the  second  formula  for  S,  we  have 
iS'- 12.  7  +  ^^^1^  =  12.7 +  12.  11  =  12(7  +  11)  = 
12  .  18  =  216. 

N.  B.  Thetwoformul8B,/=a±(w— l)rf,  and>S^=^^^i^\ 

should  be  retained  in  memory  by  the  learner. 

Art.  14:3.  The  first  and  last  terms  of  a  progression  are  called 
the  extremes ;  and,  when  the  number  of  terms  is  odd,  the  middle 


2JJ2  PROGRESSION    BY    DIFFERENCE.  J. 

one  is  called  the  mean,  but  when  the  number  of  terms  is  even 
the  two  situated  midway  between  the  extremes,  are  called  the 
means. 

Ifwe  observe  the  process  ofadding  equations  (I)  and  (2),  in  Art 
141,  it  will  be  manifest,  that  the  sum  of  any  two  terms  equally 
distant  from  the  extremes ,  is  the  same  as  the  sum  of  the  extremes 

Moreover,  if  the  number  of  terms  is  odd,  the  sum  of  the  extremes 
xaill  be  equal  to  twice  the  mean.  For,  the  number  of  terms  being 
odd,  the  middle  terms  of  equations  (1 )  and  (2)  will  be  of  the  same 
value,  although  expressed  in  one  by  a  plus  a  certain  number  of 
times  the  difference,  and  in  the  other  by  /  minus  the  same  num- 
ber of  times  the  difference.  These  two  middle  terms  therefore 
being  added,  their  sum  will  be  the  same  as  twice  one  of  them. 

Art.  143.    The  two  equations,  Izzz  a^  (n —  l)c?,  and  S z=. 
gy~    J  involve  five  different  quantities,   any  three  of  whicH 

being  given,  the  remaining  two  can  be  found. 

There  may  arise  then  the  ten  following  problems,  viz : 

1.  Given  a,  d  and  w,  find  /  and  S. 

2.  Given  a,  d  and  I,  find  n  and  S. 

3.  Given  a,  n  and  S,  find  d  and  /. 

4.  Given  a,  I  and  S,  find  d  and  n. 

5.  Given  a,  n  and  /,  find  d  and  S. 
.  6.  Given  a,  d  and  S,  find  n  and  /. 

7.  Given  d,  n  and  S,  find  a  and  /. 

8.  Given  d,  n  and  I,  find  a  and  S. 

9.  Given  w,  I  and  >S^,  find  a  and  d. 
10.  Given  d,  I  and  S,  find  a  and  n. 

The  first  of  these  problems  has  already  been  solved,  and  the 
equations  which  we  have  found  for  I  and  S,  may  be  assumed,  in 
the  solution  of  the  other  nine  problems.  Of  these  last  we  shall 
solve  the  fifth,  and  leave  the  others  to  be  performed  by  the 
learner. 

The  problem  is,  to  find  d  and  S,  when  a  I  and  n  are  known 

n(a-\-l) 
The  value  of  S  is  already  given,  viz  :  S-=z  ' — - ,  and  the 


L.  PROGRESSION    BY    DIFFERENCE.  23^3 

equation  l=a-\-{n  —  l)<f,  gives,  by  transposition  ana  livisiot. 

,       / —  a 

dz=z -. 

n  —  1 

Art.  144.  This  value  of  d  will  enable  us  to  insert  any  num- 
ber of  terms  between  two  given  quantities,  a  and  /,  so  that  the 
whole  series  shall  form  a  progression  by  difference.  The  quan- 
tities thus  inserted,  are  called  mean  differentials,  or  arithmetical 
means. 

Thus,  if  it  be  required  to  insert  m  mean  differentials  between 
the  quantities  a  and  /,  as  there  would  be  m  -j-  2  terms  in  the 
whoIe,vto  find  the  common  difference,  we  have  only  to  substitute 
m  -f-  2   instead  of  w,  in  the  formula  for  rf,  which  gives  d  = 

I  —  a  I — a 

m-\-2—i  "~  w-f-I* 

Hence, 

When  a  certain  number  of  mean  differentials  is  to  be  inserted 
between  two  quantities,  to  find  the  common  difference,  divide  the 
difference  between  the  quantities  by  a  number  greater  by  one  than 
the  number  of  terms  to  be  inserted. 

Knowing  the  common  difference,  it  is  easy  to  write  the  pro 
gression,  which,  expressed  in  general  terms,  will  be  as  follows, 
viz : 


Of 


ma-\-l  (m— l)a  +  2/   (^,— 2)a-{-3/ 
"">    m+\'  m+l  '  m+1  ' '* 

As  an  example,  let  it  be  required  to  insert  six  mean  differen 
tials  between  4  and  25. 

Here  d  =.  — r— ; .  becomes  d  =r  — - —  =  3  ;    and     he  pro 
m+ 1  7  ^ 

;?ression  is  4,  7,  10,  13,  16,  19,  22,  25. 

It  is  manifest  from  what  precedes,  that, 

20* 


234  EXAMPLES  IN  PROGRESSION  LI 

If  between  the  terms  of  a  progressitm  by  difference  taken  two 
and  twOf  the  same  number  of  mean  differentials  be  inserted,  the 
result  will  be  in  progression. 

For  example,  let  it  be  required  to  insert  between  every  two 
adjacent  terms  of  the  progression,  3,  9,  15,  21,  two  mean  differ- 
entials. 

In  this  case,  d  =  — r— -  becomes  d  z=z  — — —  =.  2 ;   and  the 
wi-j- 1  3 

orogression  is  3,  5,  7,  9,  11,  13,  15,  17,  19,  21. 


SECTION    LI. 

EXAMPLES    INVOLVING    PROGRESSION   BY    DIFFERENCK. 

Art.  14:^.  1.  How  many  strokes  does  a  clock  strike  in  12 
houf s  ? 

2.  Find  the  10th  term  and  the  sum  of  the  first  10  terms  of  the 
series,  20,  25,  30,  &c. 

3.  Find  the  16th  term  and  the  sum  of  the  first  16  terms  in  the 
series,  100,  98,  96,  &c. 

4.  Find  the  last  term  and  the  sum  of  the  series,  12, 13^,  14|, 
&c.  the  number  of  terms  being  30. 

5.  The  number  of  terms  being  28,  find  the  last  term  and  the 
sum  of  the  series,  3,  3|-,  4f ,  &,c. 

6.  Insert  six  mean  differentials  between  20  and  55. 

7.  Insert  five  mean  differentials  between  6  and  10. 

8.  Insert  five  mean  differentials  between  every  two  adjacent 
terms  of  the  progression,  5,  17,  29,  41. 

9.  Suppose  t"hat,  as  in  Venice,  a  clock  denoted,  by  the  number 
of  strokes,  the  hours  from  1  to  24,  how  many  strokes  would 
such  a  clock  strike  in  24  hours  ? 

10.  A  farmer  wished  to  set  out,  upon  a  triangular  piece  of 
^and,  25  rows  of  apple  trees,  he  first  row  containing  2  trees,  the 
second  5,  the  third  8,  and  so  on.  How  many  trees  did  he  re^ 
quire  for  his  purpose? 


LI  BT    DIFFERENCE.  235 

11  A  gardener  has  100  plants  and  a  reservoir  of  watei  all  in 
a  straight  line,  the  plants  being  3  feet  asunder,  and  the  reservoir 
10  feet  from  the  first  plant.  How  far  must  he  walk  in  order  to 
water  these  plants,  if  he  commence  at  the  reservoir,  and  return 
lo  it  for  a  new  supply  of  water  for  each  plant,  finally  coming  to 
the  reservoir  after  having  watered  the  last  one? 

12.  A  falling  body  descends,  in  vacuo,  16^  feet  the  first  sec- 
ond, and  in  each  succeeding  second,  32^  feet  more  than  in  the 
preceding.     How  far  will  a  body  descend  in  10  seconds? 

13.  We  observe  that,  in  the  preceding  question,  the  difference 
is  just  double  the  first  term.  Let  the  learner  generalize  that 
question,  by  substituting,  in  the  second  formula  for  jS^,  2  a  instead 
ofd 

14.  Two  travelers,  A  and  B,  188  miles  asunder,  set  out  at 
the  same  time  with  the  intention  of  meeting.  A  goes  regularly 
10  miles  per  day ;  but  B  goes  3  miles  the  first  day,  6  the  second, 
9  the  third,  and  so  on.     In  how  many  days  will  they  meet? 

15.  Two  men,  135  miles  asunder,  set  out  at  different  times  and 
travel  towards  each  other.  One  starts  5  days  before  the  other, 
and  goes  1  mile  the  1st  day,  2  miles  the  2d,  3  miles  the  3d,  and 
so  on.  The  other  travels  20  miles  the  1st  day,  18  the  2d,  16  the 
3d,  and  so  on.  How  many  days  and  what  distance  will  each 
have  traveled  when  they  meet  ? 

16.  Divide  51  into  three  parts,  which  shall  form  a  progression 
by  difference,  the  common  difference  being  5. 

17.  Find  three  numbers  in  arithmetical  progression,  such  tha 
their  sum  shall  be  18  and  their  continued  product  192. 

Remark.  Let  y  =:  the  common  difference,  and  x  z=  the  mea^ 
term.     Then  x  —  y,  a:,  and  x-\-i/  will  represent  the  numbers 

18.  Divide  50  into  five  parts,  which  shaJl  form  a  progression 
by  difference,  of  which  the  first  term  shall  be  to  the  last  as  7 
to  3. 

19.  There  is  a  number  consisting  of  threp  digits,  which  form 
a  decreasing  arithmetical  progression.  The  sum  of  the  digits  is 
9,  and  if  396  be  subtracted  from  the  number,  the  digits  will  be 
inverted.     Required  the  number. 


SK6  PROGRESSION    BY    QUOTIENT  LIl 


SECTION    LII. 

PROGRESSION    BY    QUOTIENT. 

Art.  14G.  A  progression  by  quotient ,  called  also  geometricai 
progression,  is  a  series  of  quantities  such,  that,  if  any  one  of 
them  be  divided  by  the  next  preceding,  the  quotient  will  be  the 
same  in  whatever  part  of  the  series  the  two  successive  terms  are 
taken. 

Progression  by  quotient  may  be  either  increasing  or  decreas 
ing.  Thus,  2,  4,  8,  16,  32  form  an  increasing,  and  60,  20,  ^^, 
^,  a  decreasing  progression  by  quotient. 

The  quotient  arising  from  the  division  of  any  term  by  that 
which  precedes  it,  is  called  the  common  ratio.  The  ratio  in  the 
first  of  the  two  progressions  given  above,  is  2,  and  that  in  the 
second  is  ^. 

In  general,  let  a,  b,  c,  d,  &.c.  be  the  successive  terms  of  a  pro- 
gression. 

Let  q  represent  the  constant  ratio ;  then,  since  each  term  is  q 
times  the  preceding,  we  have 

b=:aq,  c=iaq^,  dzzzaq^,  ez=zaq^,  &lc. 

Now  representing  the  last  term  by  Z,  and  supplying  by  points 
the  place  of  the  indefinite  number  of  terms  omitted,  the  terms  of 
the  progression  will  be, 

1st   2d    3d     4th    Sth    0th    7th 

a,  aq,  aq^,  aq^,  aq^,  aq^,  aq^, ,  /. 

We  readily  perceive,  that  the  exponent  of  q  in  any  term  is  ess 
by  unity,  than  the  number  which  marks  the  place  of  that  term. 
Thus,  the  4th  term  is  aq^,  the  Sth  aq^.  Consequently,  if  n  rep- 
resent the  number  of  terms,  the  nth  or  last  term  will  be  aq''~^. 
Therefore,  the  formula  for  the  last  term  is 

Hence. 

Any  term  of  a  progression  by  quotient ,  may  be  found,  by  mul- 
tiplying the  first  term  by  that  power  of  the  ratio,  denoted  by  a 
number  1  less  than  that  which  marks  the  place  of  the  term. 


LII.  PROGRESSION    BY    QUOTIENT.  2«i^ 

Ex.     What  is  the  sixth  term  of  the  series,  3,  6,  12,  itc? 
Here  a  =  3,  ^r  =  2,  and  nz=zQ\    therefore,  1=:  aq"'^  be- 
comes /  =  3  .  25  =  3    32  =  96. 

Ex.  2.  Required  the  7th  term  of  the  sesies  3645,  1215,  405, 
&c. 

In  this  question,  a  =  3645,  gr  =  ^,  n  =  7.  Then  /  =  3645  X 
'i)6  =  3645.^^=5. 

Art  147.  To  find  the  sum  of  any  number  of  terms,  denote 
chis  sum  by  ^S^ ;  then 

8z=.a-\-aq-\-aq^^-^aq^-{-aq^-\-aq^-\- -^  a  3"~2 -|~^  3'*~^' 

Multiplying  this  equation  by  g,  we  have 
1  Sz=zaq-{'aq^-\'aq^-\'  aq*  -\-  aq^  -\-  aq^  -{- -\-aq'^^-\-aq^ 

Subtracting  the  former  of  these  equations  from  the  latter,  ob- 
ierving  that  the  terms  of  the  second  members  all  cancel  except  a 
and  ay",  we  obtain 
^8 — Sznaq^  —  a;    or  {q  —  \)  8  z=i  a  q^  —  a;  consequently, 

_  qg"  — fl  _  «(g"— 1) 

_,       ag"  —  a         aq^^^q  —  a        ...       ,.  ,     n- 

But   — •   =  —5 ^- ;  substitutmg  /  mstead  of  lUi 

equal,  ag"~^,  we  have  Sz=:  — — . 

We  have  then  two  formulae  for  the  sum  of  a  geometrical  pro- 
gression, viz : 

a(«"  —  1)         ,    ^       ql — a      __ 

5  =  -^ — ^,  and  S  =  ± -.     Hence, 

g-— 1  q  —  l 

To  find  the  sum  of  a  progression  hy  quotient^  subtract  unity 
from  that  power  of  the  ratio ^  denoted  hy  the  number  of  terms, 
multiply  the  remainder  by  the  first  term^  and  divide  the  product 
hy  the  ratio  minus  unity  ;  or^  multiply  the  last  term  hy  the  ratio 
subtract  the  first  term  from  the  product  ^  and  divide  the  remainda 
hy  the  ratio  minus  unity. 


238  PROGRESSION    BY    QUOTIENT.  LIl 

Ex.  Required  the  sum  of  the  series,  1,  2,  4,  8,  &c.,  the 
number  of  terms  being  10  ? 

In  this  question,  a  =  1,  j  =  2,  and  n  =  10.  Therefore,  S=. 
1J21"-- 1)  _  1  (1024-1)  _ 

2_f--  -2-1        -1*^^- 

Or  we  may  first  find  the  last  term,  and  then  use  the  formula 

S=:  il^.      We   have   then   /=  1  .  29  =  512,   and   S- 

2.612-1  _1024-1_ 

2-1       -         1        -^^^' 

Ex.  2.  Required  the  sum  of  the  first  six  terms  of  the  series, 
lO,  5,  f ,  &c.  ? 

Here  a  =  10,  j  =  ^,  and  n  =  6.     Using  the  first  formula  for 

S.  we  have  ^^  lofW"-!]  ^  IQ(A-l)  ^  10  (-ff)  ^ 

Art.  14:8.  If  y  is  a  proper  fraction,  q —  1  will  be  negative, 
q^  —  1  will  also  be  negative,  since  any  power  of  a  proper  frac- 
tion, if  the  index  is  greater  than  unity,  is  less  than  the  fraction 
itself.     Changing  the  signs  of  numerator  and  denominator,  in 

1/.         1/.^         ,          «       «(l  —  Q^)         ^      ^  —  <2  7" 
the  formula  for  S,  we  have  Sz=z  —\ i-^,  or  <»=z  — ^. 

l_-.gr  1— gr 

Now,  as  the  powers  of  a  fraction  less  than  unity  become  less 
and  less,  the  greater  the  index  of  the  power,  it  follows  that  if  w, 
the  number  of  terms,  is  infinitely  great,  5"  must  be  infinte  y 

small,  that  is,  zero.     In  this  case,  Sz=z  — —  will  become  *^=x. 

1  —  q 

a  —  a  .0  a 


1-gr  l-q 

Since  q  is  supposed  to  be  a  fraction,  let  it  be  represented  b} 

~  ,  so  that  q=  —.     We  shall  then  have  o  = z=  

%*  ^       n  1 — ^       n  —  fit 


111.  PROGRESSION    BY    QUOTIENT.  239 

The  formala,  therefore,  for  the  sum  of  a  decreasing  progres 
sion  by  quotient,  continued  to  infinity,  is 


n  —  m 
Hence, 

Tojind  the  sum  of  an  infinite  decreasing  series  in  progression 
by  quotient y  multiply  the  first  term  hy  the  denominator y  and  di' 
vide  the  product  by  the  difference  between  the  denominator  ana 
numerator  of  the  ratio. 

If  9  is  a  fraction  whose  numerator  is  1,  the  formula  for  he 

sum  of  an  infinite  decreasing  series,  becomes  Sz:z -. 

n  —  1 

Ex.  What  is  the  sum  of  the  series,  5,  -^,  ^,  &c.,  continued 
lo  infinity. 

5    3 

In  this  example,  a  =  5,  and  J  =  f ;  therefore,  S  z=  = 

3 — 2 

15. 

Art.  14:9.  From  the  formula  for  the  sum  of  an  infinite  de- 
creasing progression  by  quotient,  may  be  deduced  the  rule  given 
in  arithmetic,  for  reducing  periodical  fractions,  sometimes  called 
repeating  and  circulating  decimals^  to  vulgar  fractions. 

Let  us  take  the  decimal  "333  continued  infinitely.  This  is  a 
Drogression  by  quotient,  in  which  the  first  term  a  is  -^^,  the  sec- 
ond y^^,  &/C.,  the  common  ratio  q  being  -j^.     Hence,  by  sub- 

a  n  ^      10 

stitution,   the   formula,    S  =. ,  becomes  S  =r  ^-^ == 

n — 1  10 —  I 

Again,  in  the  fraction,  '0404  &c.^  a  =.  y^^y,  and  q  z=:  ^^^ ; 
lience,  S=:  -^  becomes^^  =  ^qV—  i    =  ^• 

In  like  manner,  the  sujr    )f  -290296  &c.,  =     i^inny  ♦  1000 

1000—1 


240  PROGRESSION    BV    QUOTUNT.  LL 

Let  us  take  the  fraction  •428571428571  &lc.  Here  a  — 
NutMt^*    and  as   the  second  period    is    -000000428571,  q  =r. 

TuuifUTjTi-       Therefore  iS=: ioooooo_i ~    UUih 

which  reduced,  is  f. 

Consequently  we  see,  as  in  arithmetic,  that  the  repeating  or 
circulating  figures  are  to  he  made  the  numerator  of  a  fraction^ 
whose  denominator  is  as  many  95  as  there  are  repeating  figures^ 
and  thtn  the  resulting  fraction  is  to  be  reduced  to  its  lowest 
terms.  ■ 

If  the  repeating  figures  do  not  commence  immediately  after 
the  decimal  point,  we  have  only  to  find  the  value  of  the  repeat- 
ing part,  and  add  it  to  the  decimal  which  precedes,  reducing 
them  both  to  the  same  denominator. 

For  example,  -5333  &c.  =  -^^  -f-  y^^y  +  ^^n  &lc.,  in  which 
the  first  term  of  the  progression  is  y^^,  and  the  ratio  ^  j  in  tbis 

^    .  10 

case,  the  sum  of  the  whole  is  <^jy  -\-  ^^^  ' =  ^  -|-  f^tj    - 

Art.  l^O.    Suppose  that  a  and  /  were  given,  and  it  were  re 
quired  to  insert  between  them  any  number  of  terms,  such  tha^ 
Ihe  whole  should  form  a  progression  by  quotient.     The  term? 
thus    introduced   are   called   mean  proportionals,  or  geometric 
means  * 

From  the  equati9n  l  =  aq^—^j  in  Art.  140,  we  have 

Now,  if  it  be  required  to  insert  m  terms  between  a  and  /,  since 
there  would  be  m  -f-  2  terms  in  the  whole,  we  put  »i  +  2  instead 
of  n,  in  the  value  of  g,  just  fouiid. 


LII.  PROGRESSION    BY    QUOTIENT.  24' 

We  have  then 

m+l    _ 

q  z=  y/     -  .     Hence, 

When  any  number  of  mean  proportionals  is  to  be  inserted  he» 
ttoeen  two  quantities ^  tojind  the  common  ratio j  divide  the  greater 
quantity  by  the  less,  and  extract  the  root  of  the  quotient  to  the  de- 
gree denoted  by  the  number  of  terms  to'  be  inserted  plus  unity. 

Knowing  the  common  ratio,  it  is  easy  to  write  the  progression, 
which  is  expressed  in  general  terms  as  follows,  viz : 

m-\-l  m-j-l    OT+l    

«,«V/    i.«V/     («)•  "via)    '■ 

Ex.     Insert  five  mean  proportionals  between  2  and  128. 

OT+l 

In   this   example   the    formula,    q  =z\X    -  ,  becomes  5=. 

^JL|I  —  ^64  =  2 ;  and  the  progression  is  2,  4,  8,  16,  32,  64 
128. 

It  is  manifest,  that  the  same  number  of  mean  proportionals 
may  be  inserted  between  the  terms  of  a  progression  by  quotient 
taken  two  and  two,  and  the  result  will  be  in  progression. 

Ex.     Between  every  two  adjacent  terms  of  3,  24,  192,  insert 

two  mean  proportionals. 

3  —  3  - 

In  this  case,  q  =:  y^  =  y^S  =  2 ;  and  the  resulting  pro- 
gression is  3,  6,  12,  24,  48,  96,  192. 

Art.  151.  In  the  formulae  already  given,  a,  q  and  n  were 
■supposed  to  be  known,  and  it  was  required  to  find  /  and  S.  Bui 
if  any  three  of  the  five  quantities,  a,  q,  n,  /  and  S,  are  knowa. 
thf»  remaining  two  may  be  found. 

There  may  be,  therefore,  ten  different  problems ;  but  the  stu- 
dent, at  this  stage  of  his  progress,  is  capable  of  solving  only  four 
.>f  them.  Four  of  the  remaining  six  will  be  solved  under  thfl 
21 


242  «    EXAMPIES    IN  LIIl 

head  of  logarithms ;  but  the  two  others  give  rise  to  equations 
too  difficuh  of  solution  to  be  admitted  into  an  elementary  trea- 
tise. 

Let  the  pupil  solve  the  following  problems. 

1 .  Given  a,  q  and  n ;  find  /  and  8. 

Note,     This  question  has  already  been  solved,  and  the  results 
may  be  assumed  in  solving  those  which  follow 

2.  Given  a,  n  and  /;  find  q  and  8. 

3.  Given  q,  n  and  /;  find  a  and  8. 

4.  Given  y,  n  and  <S^:  find  a  and  /. 


SECTION    LIIl 

EXAMPLES    IJV    PROGRESSION    ¥  y    QUOTIEJVT. 

Art.  t5^.  1.  Required  the  last  te/tn  and  the  sum  of  the  pro- 
gression, 5,  10,  20,  &c.,  the  number  of  terms  being  8. 

2.  What  is  the  5th  term,  and  the  sum  of  the  first  five  terms  of 
the  progression    1,  ^,  ^,  ^c.  1 

3.  There  are  three  numbers  forming  a  geometrical  progres- 
sion, in  which  the  mean  is  10,  and  the  sum  of  the  extremes  52, 
Required  the  numbers. 

Let  x=  the  ratio.  Then  — ,  10,  and  10  x  will  represent  the 
terms 

4.  A  gentleman,  without  reflecting  upon  the  result,  agreed  to 
pay  his  gardener  1  dollar  for  the  first  month,  two  for  the  second, 
and  so  on,  doubling  his  wages  each  month  for  a  year.  What 
would  be  the  amount  of  the  year's  wages? 

5.  There  are  four  numbers  in  progression  by  quotient;  thd 
sum  of  the  first  three  is  130,  and  that  of  the  last  three  is  390 
Required  the  numbers. 

Let  X  =  the  first  number,  and  y  -=  the  common  ratio. 


'AH  PROGRESSION    BF    QUOTIENT.  243 

Then,  x,  zy,  xy^^  x  y^  will  represent  the  numbeis,  and  we 
have  the  equations, 

(1)  x  +  xy  +  xy^=:l^^; 

(2)  xy+xy^  +  xy^:=zZm, 

Sepa  ating  the  first  members  into  factors,  we  have 

;3)     ^(l+y  +  y»)=130; 
(4)     xy(I4.y  +  y2)  =  390. 

Divide  the  4th  by  the  3d ;  this  gives 

y  =  3. 

Substituting  this  value  of  y  in  the  3d,  we  have 

X  (1+3  +  9)=  130,  or 

13  X  z=  130 ;  consequently, ' 

x=10. 
The  numbers  are,  therefore,  10,  30,  90,  270. 

6.  There  are  five  numbers  in  progression  by  quotient ;  the  sum 
of  the  first  four  is  468,  and  that  of  the  last  four  is  2340.  What 
are  these  numbers  1 

7.  Divide  217  into  three  parts  which  shall  form  a  geometrical 
progression,  such  that  the  third  term  shall  exceed  the  first  by 
168. 

8.  The  sum  of  three  numbers  in  progression  by  quotient  is 
04 ;  and  the  mean  term  is  to  the  sum  of  the  extremes  a?  3  to 

10.     Required  the  numbers. 

9.  There  are  three  numbers  in  progression  by  quotient,  and 
the  sum  of  the  first  and  second  is  to  the  sum  of  the  second  and 
third  as  1  to  2.     What  are  these  numbers? 

10.  There  are  three  numbers  in  progression  by  difference, 
such  that  if  the  second  power  of  the  first  be  increased  by  1 ,  that 
of  the  second  by  5,  and  that  of  the  third  by  41,  the  results  will 
form  a  progression  by  quotient,  in  which  the  sum  of  the  terms 
will  be  130,  and  the  sum  of  the  extremes  will  be  to  the  mean  aa 
10  to  3.     Required  the  numbers. 

11.  Find  a  mean  proportional  between  9  and  4. 

12.  Find  a  mean  prooortional  between  4  and  25. 


Ji44  EXERCISES    IN    EQUATIONS  LIV 

13.  Find  a  mean  proportional  between  7  and  9,  carried  tc 
three  decimals. 

14.  Find  a  mean  proportional  between  "JS  fnd  425.  accurate 
to  three  decimals. 

15.  Find  the  sum  of  the  series,  1,  ^,  ^,  &.c.,  carried  to  in- 
finity. 

16.  Find  the  sum  of  the  series,  4,  |,  |,  &c.,  continued  to  in* 
Gnity. 

17.  Find  the  sum  of  7,  ^^j  ^,  &c.,  continued  to  infinity. 

18.  What  is  the  sum  of  81,  9,  1,  ^,  &c.,  continued  to  in- 
finity ? 

19.  Insert  three  mean  proportionals  between  2  and  162. 

20.  Insert  two  mean  proportionals  between  5  and  1080. 

21.  Insert  a  mean  proportional  between  every  two  adjacent 
terms  of  the  progression,  3,  75,  1875,  46875. 

22.  Insert  two  mean  proportionals  between  every  two  adja- 
cent terms  of  the  series,  1,  8,  64,  512. 


SECTION    LIV. 

EXERCISES    IJV    EQUATIONS    OF    THE    SECOIirD    DilOREE. 

Art.  lo3.    Solve  the  following  equations. 


1,  Given  \/6  +  3  x  =  6 ;  to  find  x. 
Squaring  both  members,  we  have 

6  +  3  X  z=  36 ;  hence, 
X  z=  10.     Ans. 

2.  Given  (16  +  x2)^  _  X  zn  2 ;  to  find  X. 
By  transposition, 

( 16  +  ^^)    =  2;  -|-  2 ;  squaring  both  members, 
1 6  -j-  x^  =  x^  -f-  4  X  4"  4 ;  transposing, 
z-  —  x2  —  4xr=^4  —  16;  reducing,  changing  «he  signs 
%nd  dividing, 

^       X  =  3.     Ans 


LIV  OF  THE  SECOND  DEGREE.  ^fl 

^J  +  28       v/^+38 

3.  Given     ,--  ,    ^    =:  --= ;  to  find  x. 

^x  +  4        v^x  +  e 

Clearing  the  equation  of  fractions  and  reducing, 

X  +  34  ^7  +  168  =  2  +  42  ^7  +  152 ;  transposing, 
X  +  34  ^7  —  X  —  42  y/z  =  152  —  168 ;  reducing, 
—  8  y/z  r=:  —  16;  changing  the  signs  and  dividing  by  8, 
^x  =  2 ;  squaring, 
a;  =  4.     Ans. 

2»n. 


4.  Given  ^x^  +  5  a  x  -|-  6^  ^^  ^a  -j-  x ;  to  find  x. 

Raising  both  members  to  the  with  power, 

^z^  -\- o  a  X -\- b'^  =:z  a -\-  z ;  squaring, 
r^ -{- 5 a X -\- b^  =.  a^ -\-2  a X  -\-  x^ ;  transposing  and  re- 
ducing, 

3  a  X  =  a2 —  59 .  dividing  by  3  a, 
a2— -62 

X  =  — .     Ans. 

oa 

5.  Given  (x  +  6)^  =  (x  — -  6)^ ;  to  find  x. 

Raising  both  members  to  the  4th  power, 
x  +  6=:(x  — 6)2,  or 

X  -|-  6  =:  x2  —  12  X  +  36 ;  inverting  the  members, 
x2  —  12  X  -j-  '^^  ^=  •*'  ~h  ^  f  transposing  and  reducing, 
x2— 13x  =  — 30. 

Now;  by  substituting  in  formula  4th,  Art.  96, 

X  =  y»  rh  ^^—^0  +  If 3-  :=  V^  i:  J.     Hence, 
X  =  10,  or  X  =  3.     Ans. 

We  see,  from  the  preceding  examples,  that  an  equation  con- 
taining radical  quantities,  may  generally  be  freed  from  them,  by 
raising  both  members  to  the  power  denoted  by  the  degree  of  the 
radical,  the  operation  being  repeated,  if  necessary.  When  some 
of  the  term?  do  not  contain  radicals,  it  is  usually  best,  in  the  first 
place,  to  make  thom  constitute  one  member,  and  the  remaining 
terms  the  other. 

21* 


i46  EXERCISES    IN    EQUATIONS  LIV 

Many  oi  the  problems  in  this  and  the  following  section  wiL 
<\ve  several  answers  each,  if  the  double  sign  dc  be  prefixed  to 
t»ots  of  an  even  degree. 

Find  X  in  the  following  equations. 

6.  ^nri  +4  =  9. 

7.  2:  +  (x  +  6)^=2  +  3(a;  +  6)^. 

8.  ^12+^=:^x  +2. 

9.  ^^  — 2  =  4— 3^£" 

10.  v/H^  =  v/^  +  1 

11.  (x  — 32)^=  16 -—2;^. 

The  mode  pursued  in  the  preceding  questions,  frequently  leads 
to  equations  of  so  elevated  a  degree,  that  their  solution  would  be 
too  difficult  for  an  elementary  work.  Other  expedients,  there- 
fore, are  often  to  be  preferred. 

Whenever  an  equation  can  be  reduced  to  the  form  of  z^"*-4- 
p  z""  :=  i  5,  that  is,  to  an  equation,  in  which  the  unknown  quan- 
tity is  found  in  two  terms  only,  and  has  an  exponent  in  one  of 
them  double  its  exponent  in  the  other,  it  may  be  solved  after  the 
manner  of  affected  equations  of  the  second  degree. 

12.  Given  x4 -1-6x2  =  135;  to  find  x. 

First  consider  x^  as  the  unknown  quantity  and  make  the  first 
member  a  perfect  square, 

X*  +  6  x2  -j-  9  =  144 ;  extracting  the  square  root, 
x2  -|-  3  =^  zh  12 ;-  transposing  and  reducing, 
x^  =:  9  or  —  15 ;  taking  the  square  root  of  this, 

X  =  rh  3,  or  X  =  rb  /— 15. 


Hence,  x  =  3,  x  =  —  3,  x  =  ^ —  15,  or  x  =:  — y/ —  15. 
This  question  might  have  been  solved  by  means  of  formula 
Is:.  Art.  96. 


Thus,  x2=: —3^^/135+ 3^6  ___3_i-i2-:9  or  =  —  15 
ihen  X  izr  ±  3,  or  X  =  ±  ^ —  15. 


LIV  OF  "^'lE  SECOND  DEGREE.  241 

13,  f^iven  a;  +  ^  i/^  =  32 ;  or  what  is  the  same,  x  +  4  r*  — 
32 ;  to  find  x. 

In  this  equation  consider  yjx  as  the  unknown  quantity.     By 
rhc  formula  1st,  Art.  96,  we  have 


^x=— 2±^32  +  4; 

^x  ■=.  4,  or  y/x  =  — 8j  squaring  both  equations, 
X  =r  16,  or  z  =  64.     Ans. 


4  — 


14.  Given  2  ^z  +  3  ^'x  =  27 ;  to  find  x. 
Dividing  by  2, 

Referring  to  the  formula,  considering  i/x    as  the  unknown 


quantity, 

4 


\/x  =  — |±^y+Ai  hence, 

l^x  =  3,  or  ^x  =  —  f ;  taking  the  4th  power  of  both 
equations, 

x=:81,  or  x=  ^fJ-.     Ans. 

15.  Given  x*  +  10  =  5  x^  +  4  ;  to  find  x. 
Transposing  and  reducing, 

X* — 5  X*  =  —  6 ;  completing  the  square, 

X* — 5x*-|-  ^  =  — 6+  ^  ;   or,  reducing  the  second 
member, 

X*  —  5  X*  -|-  \*  =  i  J  taking  the  square  root, 

x^  —  |-  =  rh  ^ ;  transposing  and  reducing, 

X*  =  3,  or  I*  rr  2 ;    raising  both   equations   to   the   6th 
power, 

x  =  729,  or  x  =  64.     Ans. 

If  we  had  substituted  in  formula  4th,  Art.  96,  the  operation 
ivould  have  been  shorter. 


248  EXERCISES    IN    EQUATION^  i'  l'^ 

16.  Given  y/x^  +  ^x^  =z  6  ^x ;  to  And  x. 
Taking  the  roots  of  the  perfect  squares  and  piaoing  tntm  b 
fore  the  radical  sign, 

x^  i/x  -f-  ^  1/2;  =  G  i/x  ;  dividing  by  i/i, 
x^-\-x  =  6. 
This  may  now  be  solved  like  any  affected  equation ;  and    he 
following  equations  may  be  solved  like  the  preceding. 

17.  2x4  +  8x2  =  24. 

18.  x6_8x3  — 513  =  0. 


10 

3^x                x_5_ 

5                     20    - 

20. 

x^  +  7x^  —  U  =  0. 

21. 

4  x^  +  x^  =  39. 

22. 

3x6  +  42x3  =  3321. 

23. 

x^+x^  =  6x^. 

24. 

x_4x^  =  45. 

25. 

4x*  =  7x^  — 6. 

26.  5x*  — 3x^  =  4x^  +  342. 

27.  3  X  —  4  ^x  =  240. 

28.  3y/r +  7^7  =  48. 

29.  V^^  +  ^  =  lHVA. 


4  +  ^x  ^x 

30.  y/x^  —  2^x  -—  X  =  0. 

31.  ax^  +  Ax^  =  c. 


32.    Givenx  +  5=^x+5  +  6;  to  find  x. 
By  transposition, 

x+5— ^^+5  =  6. 

Considering  ^x  +  5  as  the  unknown  quantity,  and  completing 
the  square, 


LIV  OF    THE    SKCUiNU    DEGREE.  249 


a,.-|-5  — ^x-f  5  +  ^  =  6  +  i=¥;  making  the  rcx>t 


tions, 


i/^  -j-  5  —  ^  :=  rb  I ;    ransposing  and  reducing, 

i/i -[- 5  =  3,  or  i/a;  +  5  =  — 2;    squaring  both  equa- 

x-(-5=:9,  orx-|-5  =  4i  transposing  and  reducing, 
=1 4,  or  X  =  —  1.    Ans. 

Another  method. 


Resume  the  equation,  x-\-^  —  x/x  -\-^:=zQ. 


Substitute  some  letter,  as  y,  instead  of  ^x+^5  ^^^^  we  have 

^x  -|-  5  =  y,  and  consequently  x  -)-  5  =  v'^ ;  hence,  the 
eq'iation  becomes 

y^  — y  =  6.     This  gives,  by  the  formula, 

y  =  ^±  v/6  +  1  =  3,  or  y  =  —  2 ;  therefore, 
y2  =  9;  ory^  —  A. 

But  y'^z=.x-\-^\  hence, 

x4-5  =  9;  orx-|-5=:4;  transposing  and  reducmg, 
x  =  4t;  orxz=:  —  1.     Ans. 

The  latter  method  of  solution  is  preferable,  as  it  saves  the  ne« 
cessity  of  repeatedly  writing  a  polynomial. 

33.  Given2x2  +  3x  — 5^2x24-3x4-9  =  — 3;  to  find  z 

Adding  9  to  each  member. 


2x2  4-3x4-9  — 5y/2x2 4-3x4- 9  =  6. 
Lety  =  ^2x2  4-3x  +  9;  then  y2  — 2x2  4-3x4-9;  hence 
y2 — 5y  =  6;  then,  by  formula  2d,  Art.  96, 


y  =  ^=tv/6  +  V==6,oryr=  — I;  hence, 
y^  =  36,  or  y2  _  1,     Taking  the  1st   -alue  of  y^ 
2x94-3x4-9rr  36; 
x2  4-|x  =  V| 

z  =  3 ;  or  X  =  —  |. 


250  EXERCISES    IN    EQUATIONS  L\ 

Taking  the  other  value  of  y^^  viz  :  1, 

—  3d=v/— 55 

-= — t — • 

Solve  the  following  equations. 


34.  2  +  16  — 7^z  +  16=:10  — 4^x  +  16 

35.  x  +  ^i+r6=:2  +  3y/x  +  6. 

36.  x2  — 2z  +  6^x2— 2ar  +  5  =  ll. 

37.  (lO  +  a;)^— (10  +  a:)^  =  2. 

38.  (a;— 5)3— 3(2— 5)^  =  40. 


v/a:24.^_|,6        18- (|v/x2  + 2  +  6-^^2) 

3  -  ^2:2  +  2  +  6 

40.   fl;2_a;_^5^2x2— 52  +  6  =  ^"'^^. 


SECTION    LV. 

CXERCISKS    IN    EQI7ATION8    OF    THX    SECOND    DEGREE   WITH   TWO    tJlT 
KNOWN   QUANTITIES. 

Art  154.     1.  Given  ^  ""J^  7-^1^^^  \  '  *^  ^^^  *  ^""^  ^ 

From  the  1st  equation, 

X  =:2y ;  substitute  this  value  of  y  in  the  2d  equaticn, 

4y2_y2  =  i2; 

3y2=,i2; 

y2  =  4; 

y  =  ±2. 


LV  OP  THE  SECOND  DEGREE.  251 

Hence,  x  =:  2  y  =  ±  4. 

In  the  preceding  question,  the  value  of  one  unknown  quantity 
was  found  in  one  equation,  and  substituted  in  the  other ;  and  in 
this  way  the  solution  can  be  effected,  when  one  of  the  given 
equations  is  of  the  first  degree,  and  the  other  of  the  second. 

Find  X  and  y  in  the  following  equations. 

42^  +  23^ 
2. 


5. 


2y  +  4z=:ll. 


+  33^^ 

When  both  equations  are  above  the  first  degree,  different  ex 
pedients  are  to  be  adopted,  which  will  be  best  learned  from 
examples. 

((1)     zy  =  50. 

Adding  twice  the  1st  to  the  2d, 

(3)  x2  -f-  2  z  y  4"  J^^  =  225 ;  taking  the  square  root 

(4)  x  +  y  =  ±:15. 

Su  )tr acting  twice  the  1st  from  the  2d, 

(5)  x2  —  2  X  y  +  y2 ::-  25 .  taking  the  square  root, 

(6)  X  — y  =  ±:5. 

Adding  tho  4th  and  6th,  and  dividing  by  2, 

x  =  ±10. 
Subtracting  the  6th  from  the  4th,  and  divid  ng  by  2, 

3^==h5. 


252  EXERCISES      N    EQUATIONS  L\ 

By  taking  all  the  possible  combinations  of  the  signs  in  the  2d 
members  of  the  4th  and  6th,  we  have 

X  =  10,  and  y  =  5 ;  or, 
x=z  — 10,  and  y  =:  —  5  ;  c  r, 
z  =  5,  and  yzz:  10;  or, 
X  =  — 5,  andy  =  — 10. 


5(1)     z^+xy  =  li. 
h2)     xy  +  y^  =  2i. 


Aoding  the  1st  and  2d, 

(3)  x2  _j_  o  3.  y  _^  y2  _.  36 .  taking  the  root. 

(4)  x  +  y  =  ±6. 

But,  x^-\-xi/=  12  is  the  same  as 

{x-\-7/)x=i  12;  substituting  in  this  dt  6  instead  of 

=b6x=12; 

x  =  db2. 

Substituting  this  value  of  x  in  the  4th, 
db2+yr=d;:6; 
i/  =  ri=6iF2,or, 
i/  =  rt  4. 

In  the  last  equation  but  one,  the  upper  signs  correspond,  as 
also  do  the  lower. 

((1)     x  +  y  =  s. 

Squaring  the  1st, 

(3)  x^-\-^xy-\-y^z=:s^\   subtracting  from  this  4 
rimes  the  2d, 

(4)  X*  —  2xy-|-y^  =  ^'^  —  'ia^'f  taking  the  square 
root. 


(5)     X  — y  —  dcv/^'^  — 4a2. 
Adding  the  1st  and  5th,  and  dividing  by  2, 
s-j=i\/s'^  —  4a^ 

V  —  !1 


LV.  •  O/    THE    SECOND    DEtJREE.  25o 

Subtracting  the  5th  from  the  1st,  and  d.viding  by  2, 


2 

l('Z)     2:3 +  2/3 —189. 

Adding  3  times  the  1st  to  the  2d, 

(3)  23  -^  3  x2  y  +  3  X  y2  _|_  y3  _  709  ^  taking  the  3d 
<oot, 

(4)  x  +  y  =  9. 

But  the  first  member  of  the  1st  is  the  same  as  xi/{x-\-i/)  : 
substituting  9  for  x  +  y,  the  1st  becomes 

(5)  9x^=180;  hence, 

(6)  xy  =  20. 
Squaring  the  4th, 

(7)  x^ -|-2  xy -|-y^  =  81  ;    subtracting  from  this  4 
limes  the  6th, 

(8)  x2  —  2  X  3^  +  y9  :=  1  ;  taking  the  2d  root, 

(9)  X  —  y  =  dc  1 ;  adding  the  4th  and  9th  and  divid- 
ing by  2, 

0  z=  — - —  2=z  5,  or  4. 

Subtracting  the  9th  from  the  4th,  and  dividing  by  2, 


=  4,  or  5. 


(1)     x4-y4^i776. 
^(2)     x2  — y2_24. 


Di\ide  the  1st  by  the  2d, 

(3)     x2  +  y2  _  74  .  adding  the  2d  and  3d,  and  divid- 
ing  b}  2^ 

x2  :=  49 ;  hence, 
x  =  db7. 
Subtracting  the  2d  from  the  S-*.  in  ^  divf'mjr  by  2, 
w2z=25;  henc«,   ^ 

y  =  ±5. 

22 


254  EXERCISES    IN    EQL  VilONS  LV 

^^       C(l)     z2  +  x  +  yz=  18-3/9 

From  the  1st  by  transposition, 

(li)     x^ -\- X -\- y -\- y^  z=z  \S  \  adding  to  this  twice  the  2d, 

(4)  2;2  _|_  2  X y -J- y2  _j_ 3;  _j_ y  — -  30 ;  or  what  is  the  same, 

(5)  (z  +  y)^  +  (^,H~  y )  ^^^  *^^  5  completing  the  square, 

(6)  (a:  +  y)2  +  (a:  +  y)  +  ^  =  30  +  i=-L2i.    taking  the 

2d  root, 

(7)  a:  +  y_(-^  — -t-^;  whence, 

(8)  x  +  y  =  5,or-6. 

Substituting  these  values.of  x -f-y  in  the  3d,  and  transposing^ 

(9)  x2  -|- y2  _-  13^  Qf  24 .  fi-ojjj  ti^ig  subtracting  twice  the  2d, 

(10)  x2  — 2xy  +  3^2_i^  or  12;  taking  the  2d  root 

(11)  X  — 3/  =  ±l,  or  d=2^37  adding  the 8th  and  11th,  and 
dividing  by  2, 

x  =  3,  or2,  or — 3-fcy/3;   subtracting  the  llth  from 
%h,  and  dividing  by  2, 

y=:2,  or  3,  or  — 3=Fv/3r 

Solve  the  following  equations. 

y  — 3y 
18. 


12 
13. 
14 


^X2/=Z 

Cx  +  y:::zlO. 


y  =  21. 
15. 


\xy=z 
jg     Cx2  — xy  =  54. 
.  *     \xy  —  y^=\ri 

16  fv/f;+v/^=i2' 

17.  J^^-y*~=-* 

U* +  3/^  =  7. 


LV  OF    THE    SECOND    DEGREE.  25A 

18. 

/'    x-4-  V 


2:4 —  3^4 -.369. 


19.    Zx^  —  y^ 

/2 


20 
21. 


4zy=:96  — x2y2. 
X  +  3^  =  6. 
Remark.     Find  the  value  of  xy  in  the  Ist,  as  an  affected 


equation. 


Remark.     Divide  the  2d  by  the  1st. 


22     52;2y  +  xy9  =  6. 

^a:3  3/2_^x2y3--i2. 

23.    ^^+V/^  +  i^=19- 


^'  +  ^y  +  y^  =  84. 

Sometimes,  when  an  equation  is  given  in  the  form  of  a  pro- 
portion, it  may  be  transformed  to  a  more  simple  proportion,  and 
the  solution  may  be  facilitated  by  means  of  the  principles  given 
in  Art  139. 

Find  the  values  of  x  and  y  in  the  following  examples. 

2=:     5x+y:x  — y=l3:5. 

^  •     tx  +  y^  =  25. 
The  1st  gives,  by  Proportions,  11th, 

2  X  :  2  y  =  18  : 8 ;  hence,  Prop.  6th 
X  :  y  =  9  :  4 ;  and.  Prop.  3d, 

Substituting  this  value  of  x  m  equation  2d, 
y^ -{- -J- =:  25 ;  whence. 


2a6  EXFKcisr.s  fn  f.qijations  1  V 


y  =z  4,  or  —  G^.     Cousequeiitly, 
1  =  ^  =  9,  or- 14,V 


\  (2)      x3 «3  .  (^.  _  „^3 


=  24. 

Dividing  both  terms  of  the  1st  ratio  of  2d    jy  2; —  y,  Prop 
6lh, 

(3)  x2-f  2y-fy2:(._y)-2^19.1,or, 

(4)  x2  +  xj/-{-y9.  2;2_2zy  +  3^2_  j9.  J  .  hence 

Hop.  lOth, 

(5)  3  2;  y  :  18  =:  3:2  _  2  J  y  -I-  y2  .  1  .    substituting  24 
for  xy  in  the  1st  term, 

(6)  72  :  18=  (x  — 3^)2  :  1  ;  hence,  Prop.  6th, 

(7)  4  :  1  =  (x  — 3/)2  :  1 ;  and,  Prop.  14th, 

(8)  (x  —  y)2  =  4  ;  from  which, 

(9)  X— y  =  ±2. 

Adding  4  times  the  1st  to  the  square  of  the  9th, 

(10)  x2  -f  2  X  y  -f  y-  =  1 00 ;  extracting  the  root, 

(11)  x  +  y  =  =hlO. 

Adding  the  9th  to  the  11th,  and  dividing  by  2, 

z  =  rt  6. 
Subtracting  the  9th  from  the  11th,  and  dividing  by  2, 

y  =  =h4. 


27   5^-.y  =  ^^4. 

^x+5:i/— 1=5:3. 


28. 


X  :  ;/  =  5  :  3. 


29      5-^-48. 


y3_y3..  (:^_y)3  — 37-1 

Cx2:y2:,^49:36. 
i2x-y;x  +  6=16.i?0. 
«,      Jy.So  — y  =  x:10  — X. 
^2y— 41  =  (11— x)2. 


LV  OF    THE    SECOND    DEGREE  257 


x2-(-y2.^2_^9_  17:8. 
:y2  — 45. 

When  two  equations  of  the  second  degree,  containing  two 
unknown  quantities,  are  homogeneous  with  respect  to  these 
quantities,  that  is,  when  each  unknown  term  contains  either  the 
square  of  one  of  the  unknown  quantities,  or  the  product  of  both, 
the  solution  can  be  effected  by  substituting,  for  one  of  the  un- 
known quantities,  the  product  of  the  other  by  a  n'^w  unknown 
quantity. 

C(l)    2»^-iy  =  6. 

Let  x=zzy. 

Substituting  this  value  of  x  in  the  given  equations, 

(3)  ^z^y^  —  zy^=z^. 

(4)  2 3^2 _|_ 3 2^2 —a     From  the  3d, 

(^)     ^^  =  2T3-z      "^"^^' 

Solving  this  equation, 


2  +  32         222__z- 


hence. 


^=  if  zbv/f  +  (if)"  =  if  db  v/Mi  =  2,  OF 

8 


y«  = 

"2  +  6 

=  1 ;  and, 

y  = 

:  ±  1 ;  or. 

y^-- 

8 
"2-f 

=  ^;  and. 

y  = 

-c*- 

Therefore, 

x  =  zy 

=  dbl  .2  = 

:±2 

:  or, 

X  = 

(-1). 

92* 

(-1^)= 

=  ^ 

3 

25S 


LOGARITHMS. 


LVl 


Solve  the  following  equations  in  a  similar  m  jiner. 


34. 


35. 


36 


37, 


<x^-{-xy=i  12. 

y6x^  —  ^xy-{-y^  =  2l. 
\2xy  =  ^y^-\.x^-^\9 

<  4  a;2  =  32:5^  —  2. 
^  x2  +  y2  _  5. 

5  x2  _  3  X  y  z^  56. 
^y^  +  xy  =  %S. 


SECTION    LVI. 


LOGARITHMS. 


Art.  1^^.  We  have  already  seen,  that  two  different  powers 
of  the  same  quantity  are  multiplied  together  by  adding  the  expo- 
nents, and  divided  one  by  the  other  by  subtracting  the  exponent 
of  the  divisor  from  that  of  the  dividend ;  also,  that  any  power  of 
a  quantity  is  found  by  multiplying  the  exponent,  and  any  root  is 
found  by  dividing  the  exponent,  by  the  number  expressing  the 
degree  of  the  power  or  root. 

Let  us  construct  a  table  of  powers  of  any  number,  as  2,  for 
example,  placing  the  powers  in  one  column  and  the  exponents 
'id  another. 

Exponents. 
16 
17 
18 
19 
20 
21 
22 
23 


owers. 

1 

Exponents. 
0 

Powers. 
256 

Exponents. 

8 

Powers. 
65536 

2 

1 

512 

9 

131072 

4 

2 

1024 

10 

262144 

8 

3 

2048 

H 

524288 

16 

4 

4096 

12 

1048576 

32 

5 

8192 

13 

2097152 

64 

6 

16384 

14 

4194304 

>«xa 

7 

32768 

15 

8388608 

LVI  *  lOGARITllMS.  25*J 

Suppose  now  it  were  required  to  multiply  1^28  by  1024. 

Looking  in  the  table,  we  find  against  128  the  exponent  7,  and 
against  1024,  the  exponent  10;  these  exponents  being  added  give 
17.  We  now  find  17  in  the  column  of  exponents,  and  against 
it,  in  the  column  of  powers,  we  find  131072,  which  is  the  pro- 
duct of  128  by  1024 ;  that  is,  128  .  1024  =  2^  .  2^^  =z  2^^  =z 
131072. 

Divide  2097152  by  64. 

Looking  iff  the  table,  we  find  21  for  the  exponent  correspond- 
ing to  the  dividend,  and  6  for  that  corresponding  to  the  divi- 
sor ;  subtracting  the  latter  from  the  former,  we  have  15  for  the 
exponent  corresponding  to  the  quotient;  we  now  find  15  among 
the  exponents,  amd  against  it  stands  32768,  which  is  the  quotient 

2097152        221 
required.     That  is,      J^       ='^=2'^  =  32768. 

Find  the  third  power  of  64 

The  exponent  against  64  is  6,  which  multiplied  by  3  gives  18 ; 
against  the  exponent  18  we  find  262144,  which  is  the  power  re- 
quired.    That  is,  (64)3  _  (26)3  _  218  ~  262144. 

Find  the  fifth  root  of  32768. 

Against  32768  we  find  the  exponent  15,  which  divided  by  5, 
gives  3 ;  against  the  exponent  3  stands  8,  which  is  the  root  re- 
quired.    That  is,  (32768)^  =  (2^5)i  =  2^  zn  2^  =  8. 

Let  the  learner  find  the  answers  to  the  following  questions  by 
means  of  the  table. 

1.  Multiply  16  by  128. 

2.  Multiply  1024  by  64. 

3.  Multiply  512  by  2048. 

4.  Multiply  256  by  16384 

5.  Multiply  256  by  512. 

6.  Multiply  32768  by  128. 

7.  Divide     2097152  by  65536 

8.  Divide     32768  by  1024. 

9.  Divide     262144  by  16384. 


SWM)  LOGARITHMS  .  LVl 

10    Divide     524288     by  512. 

11.  Divide     4096         by  256. 

12.  Divide     8:388608  by  131072. 
13    Find  the  3d  power  of  32. 

14.  Find  the  2d  power  of  128. 

15.  Find  the  4th  power  of  16. 

16.  Find  the  2d  powder  of  1024. 

17.  Find  the  4th  power  of  32. 

18.  Find  the  5th  power  of  16. 

19.  Find  the  2d  root  of    1024. 

20.  Find  the  3d  root  of    512. 

21.  Find  the  6th  root  of    262144. 

22.  Find  the  4th  root  of    65536. 

23.  Find  the  5th  root  of    32768. 

24.  Find  the  7th  root  of    2097152. 

Art  1^6.  The  number  2,  which  is  raised  to  the  several  pow- 
ers in  the  preceding  table,  is  called  the  base  of  the  table ;  and 
the  exponents  of  the  several  powers,  are  called  logarithms  of  the 
numbers  to  which  those  powers  are  equal.  Thus,  if  2  is  the  base 
of  the  table,  the  logarithm  of  256  would  be  8,  and  that  of  16384 
would  be  14. 

A  table  might  be  made,  having  for  its  base  3,  5,  or  any  num- 
ber except  1.  Unity  would  not  answer  for  a  base,  because  all 
the  powers  as  well  as  all  the  roots  of  1  are  1. 

Tables  of  logarithms  in  common  use,  are  constructed  upon 
the  number  10  as  a  base. 

Hence, 

The  common  logarithm  of  a  number,  is  the  exponent  of  tht 
power  to  which  10  must  be  raised,  in  order  to  produce  that  num* 
her. 

Thus,  3  is  the  logarithm  of  1000,  because  10^  =  1000 ,  and 
0 5  is  the  logarithm  of  3162277,  because  lO^s  =  ^10  = 
5  162277,  nearly. 

Remark.  We  shall  hereafter  use  the  expression  log  for  the 
words  "logarithm  of." 


LVI.  LOGARITHMS.  261 

Now  100  --  1^  10'  =  10,  102  _  100,  103  —  1000,  10*  =l 
lOOOO,  &LC.  Therefore,  log.  1  =  0,  log.  10  =  1,  log.  100  =  2 
log.  1000  =  3,  log.  10000  =  4,  &c. 

Again,  10-i  =  tV  =  -1,  10-2==^^^  = -01,  10-3  =  .^^^^^  = 
•001,  10-4  =  ^^^=  -0001,  Art.  133.  Therefore,  log.  -1  = 
—  1,  log.  -01  =  — 2,  log.  -001  =  — 3,  log.   0001=  — 4. 

Hence,  the  logarithm  of  a  number  between  1  and  10  must  be 
a  fraction,  that  of  a  number  between  10  and  100,  1  +  a  frac- 
tion, that  of  a  number  between  100  and  1000,  2  -}-  a  fraction, 
and  so  on 

It  also  appears,  that  the  logarithm-  of  a  fraction  less  than 
unity,  must  be  negative,  and  that  the  logarithms  of  intermediate 
numbers  between  '1  and  01,  01  and  '001,  -001  and  0001,  will 
consist  of  whole  numbers  and  fractions. 

Art.  1^7.  To  form  a  conception  of  the  construction  of  loga- 
rithms, let  us  place  some  of  the  powers  of  10  in  one  line,  and 
their  exponents  or  logarithms  in  another  beneath.     Thus, 

1  or  100,     10,     100,     1000,     10000,     100000. 
0,  1,       2,         3,  4,  5. 

If  we  examine  these  two  series,  we  shall  perceive  that  the  for- 
mer is  a  progression  by  quotient,  and  the  latter  a  progression  by 
difference.  . 

Now  if  we  insert  between  the  terms  of  the  former  series  taken 
two  and  two,  any  number  of  mean  proportionals.  Art.  l«iO,  and 
between  the  terms  of  the  latter  taken  two  and  two,  the  same 
number  of  mean  differentials.  Art.  14:4r,  the  terms  of  the  latter 
result  will  be  the  logarithms  of  the  corresponding  terms  of  the 
former  result. 


Thus,  the  mean  proportional  between  I  and  10  =z  i/l  .  10  = 
8  162277  ;  and  the  mean  differential  between  0  and  1  ^z -L^  zzr 
j  —  -5.     Then,  log.' 3' 162277  =  5. 

If  however  we  were  to  insert  a  very  great  number  of  mean 
proportionals  between  1  and  10,  we  should  find  among  them 
terms  which  differ  very  little  from  2,  3,  4,  5,  6,  7,  8  and  9,  and 
ivhich  therefore  might  be  considered  equa    to  these  numbers 


i^'-l  LV.GARITHMS.  LVl 

Indeed  this  difference  would  be  less,  in  proportion  as  the  numbei 
of  means  inserted  was  greater,  so  that  the  approximation  might 
be  carried  to  any  degree  of  accuracy. 

If  we  insert  now  between  0  and  1  a  number  of  mean  differen- 
tials, equal  to  the  number  of  mean  proportionals  inserted  be- 
tween 1  and  10,  the  terms  of  the  result  would  be  the  logarithms 
of  the  terms  of  the  series  previously  found,  and  those  correspond- 
ing to  2,  3,  4,  &c.  would  be  the  logarithms  of  these  numbers. 

This  process  which  we  have  given,  is  designed  to  show  the 
learner  the  possibility  of  constructing  logarithms,  rather  than  aa 
a  mode  which  can  conveniently  be  reduced  to  practice. 

The  methods  by  which  logarithms  are  actually  calculated,  are 
in  general  very  different  from  that  given  above,  and  are  too  com- 
plicated to  be  introduced  into  an  elementary  treatise. 

Suppose  then  that  we  have  a  table  containing  the  logarithms 
of  all  the  natural  numbers,  1,  2,  3,  &/C.,  to  any  definite  extent, 
[n  such  a  table  the  logarithm  of  2,  for  example,  is  '30103 ;  that 

30  103^. 

is,  IQT^^^^^  =  2.  This  signifies,  that,  if  10  were  raised  to  the 
30103d  power,  and  then  the  100000th  root  were  extracted,  the 
result  would  be  very  nearly  2. 

Art.  158.  Since  logarithms  are  exponents,  they  are  subject 
to  the  rules  previously  given  for  exponents.     Hence, 

1.  To  multiply  numbers  together ^  add  their  logarithms;  the 
sum.  will  he  the  logarithm  of  the  product. 

2.  To  divide  one  number  by  another ^  subtract  the  logarithm 
^f  the  divisor  from  that  of  the.  dividend;  the  difference  will  be  the 
logarithm  of  the  quotient. 

3.  To  raise  a  number  to  any  power,  multiply  its  logarithm  by 
the  number  expressing  the  degree  of  the  power ;  the  product  will 
he  the  logarithm  of  the  power  required. 

4.  To  extract  any  roof  of  a  number ,  divide  its  logarithm  by 
the  number  expressing  the  degree  of  the  root,  or,  what  amounts  to 
the  same,  multiply  its  logarithm  by  the  fractional  exponent  which 
indicates  the  root ;  the  result  will  be  the  logarithm  of  the  root  rt 
\uired. 


'/\'\.  LOGARITHMS.  2(^3 

<>.  Since  a  fraction  expresses  division^thc  logarithm  of  afrac* 
lion  is  foundy  by  subtracting  the  logarithm  of  the  denominator 
from  that  of  the  numerator. 

6.  The  logarithm  of  either  extreme  of  a  proportion  is  found  bp 
adding  the  logarithms  of  the  means,  and  from  the  sum  subtract- 
ing the  logarithm  of  the  other  extreme  ;  also  the  logarithm  of  either 
mean  is  found,  by  subtracting  that  of  the  other  mean  from  thi 
sum  of  the  logarithms  of  the  extremes. 

Art.  loO.  In  constructing  a  table  of  logarithms,  it  is  only 
necessary,  in  the  first  place,  to  calculate  those  of  the  prime  num- 
bers ;  from  these  the  logarithms  of  all  compound  numbers  may 
o*»  deduced  by  addition  and  multiplication. 

Thus,  the  logarithms  of  2  and  3  being  found,  by  adding  them 
we  have  that  of  6.  In  fact,  the  log.  2  ==  0-3010300,  and  log.  3 
=  0-4771213;  hence  log.  6  =  03010300  +  0-4771213  = 
0-77H1513. 

Again,  2  X  log.  2  —  0-6020600  ==  log.  4,  and  3  X  log.  2  = 
0-9030900  =  log.  8,  &c. 

Hence,  from  the  logarithms  of  2  and  3,  we  easily  obtain  those 
ot  all  the  powers  oi  these  numbers,  as  well  as  those  of  all  the 
combinations  of  these  powers. 

From  the  mode  of  performing  multiplication  by  means  of  log- 
arithms, it  follows  that  the  logarithms  of  those  numbers  which 
are  10,  100,  1000,  «fec.  times  the  one  of  the  other,  must  have 
their  decimal  parts  the  same,  and  can  differ  only  with  regard  to 
their  integral  parts. 

Thus,  the  logarithm  of  2  being  03010300,  the  logarithm 
of  10  .  2  or  20  =r  log.  10  +  log.  2=1  +  0-3010300  — 
1 '3010300;  log.  200  =  log.  100  -f  log.  2  —  2-3010300;  log 
2000  —  log.  1000  +  log.  2  z=z  3-3010300.  In  like  manner,  the 
og.  3  being  0-4771213,  we  have  log.  30  =  1  4771213,  log.  30(: 
=:  2-4771213,  log.  3000  i=  3-4771213,  &c. 

Again  the  logarithm  of  371250  being  5-5696665,  we  have 
log.  37125  —  log.  (3X|2^)  =  4-5696665, 
log.  3712-5  =  log.  («-^\p)    =  3-5696665, 


LOGARITHMS 

371-25 

=  log.  (3-V^)    = 

2-5696665, 

37125 

^log.  (3VV2^)    = 

1-5696665, 

3-7125 

=  log.  (^7^^25)  •  :=, 

0-5696665, 

-37125 

=:\og.{^\h'')        = 

—  1-5696665, 

•037125 

=  log.  (HF-^)   = 

—  2-5696665, 

•0037125 

=  log.(-^A^^)  = 

—  3-5696665, 

264  LOGARITHMS  LVi 

log 

log 

log. 

log. 

log. 

log. 

log.    00037125  =  log.  (  aa3  7j^)  _  _  45696665. 
n  dividing  by  10,  in  each  instance,  we  have  subtracted  the 
.  )garithm  of  10,  which  is  1,  from  the  logarithm  of  the  dividend. 
In  the  last  four  examples,  the  subtraction  is  represented  only,  the 
decimal  part  being  positive. 

Art.  i60.  We  have  before  shown,  that  the  logarithms  ol 
fractions  less  than  unity,  are  negative;  as  represented  above, 
however,  the  integral  part  alone  is  negative.  But  the  negative 
part  being  greater  than  the  positive,  the  expression  as  a  whole  is 
still  negative.  Since  negative  logarithms  do  not  occur  in  the 
tables,  we  use  the  logarithms  of  decimals  in  the  form  given; 
and,  to  distinguish  them  from  logarithms  wholly  negative, 
we  place  the  minus  sign  over  the  integral  part.  Thus,  log. 
•0037125  =  3-5696665. 

The  integral  part  of  a  logarithm  is  called  its  characteristic, 
because  it  always  determines  the  order  of  units,  expressed  by  the 
first  significant  figure  of  the  corresponding  number. 

From  wh?»t  precedes  we  see,  that. 

The  characteristic  if  positive^  is  always  one  less  than  the  nunt' 
bet  of  integral  figures  in  the  corresponding  number ;  but,  if  the 
characteristic  is  negative,  it  is  always  equal  to  the  number  of 
places  by  which  thefrst  sigvif  cant  figure  is  removed  to  the  right 
of  the  decimal  point.  ■ 

Thus,  if  2  be  the  characteristic,  there  would  be  three  figures, 
preceding  the  decimal  point  in  the  corresponding  number ;  but 
if  the  characteristic  be  1,  the  first  figure  of  the  number  would  be 
tenths,  if  it  be  2,  the  first  figure  would  be  hundredths. 

In  logarithmic  tables  the  characteristic  is  usually  omitted. 
Bint©  we  can  never  be  at  a  loss  to  determine  it,  and  since  the 


l^Vll  UBE    OF    LOGARITHMIC    TABLES.  2C^•^ 

same  c.ecimal  part  of  a  logarithm,  correspor.  ds  to  several  difTerent 
numbers,  composed  of  the  same  figures,  but  in  which  the  figures 
express  different  orders  of  units. 


SECTFON     LVIl. 

C8E    OF    THE    TABLES    IN    FINDING    THE    LOGARITHMS    OF    GIVEN 
NUMBERS,    AND    THE    REVERSE. 

Art.  IGl.  Logarithmic  tables  are  usually  accompanied  with 
directions  for  using  them,  which  are  somewhat  different  m  aifTor- 
ent  works,  according  to  the  arrangement  and  extent  of  the  tables 
The  principle,  however,  is  in  all  cases  the  same. 

In  some  tables,  logarithms  are  carried  only  to  five,  in  others  to 
six,  and  in  others  to  seven  decimal  places. 

The  student  is  supposed  to  be  provided  with  a  table  of  loga- 
rithms carried  to  seven  decimals,  extending  to  the  number  10000. 
If,  however,  his  tables  are  carried  only  to  five  or  six  decimals,  he 
may  disregard  the  last  two,  or  the  last  decimal,  in  the  logarithms 
which  occur  below. 

Art.  lOS.  To  ^nd  from  the  tables  the  logarithm  of  a  given 
number. 

If  the  number  consists  of  less  than  four  figures,  whether  it  be 
integral,  mixed  or  decimal,  find  the  figures  in  the  left  hand  col- 
umn marked  N.  or  Number,  and,  on  the  same  horizontal  line,  in 
the  next  column  to  the  right,  will  be  found  the  decimal  part  of 
its  logarithm,  to  which  prefix  the  proper  characteristic.  In  this, 
and  all  other  cases,  zeros  to  the  right  or  left  of  the  number  will 
have  no  effect  on  the  decimal  part  of  the  logarithm.  Thus, 
log.  37  z=  ] -5682017;  log.  3700  =z  35682017;  log.  385  — 
2-5854607  ;  log.  385000  =  55854607  ;  log.  257  —  0409933 1 ; 
og.   0573  =  2  7581546. 

To  find  the  logarithm  of  a  number  consisting  of  four  figures 
ook  for  the  first  three  figures  in  the  left  hand  colamn,  md  the 
23 


J!00  •        USE    OF    LOGARlTUimC    TABLES.  LVll 

fourth  at  the  top  of  the  page ;  then,  on  the  same  horizontal  line 
with  the  first  three,  and  beneath  tiie  fourth,  that  is,  in  the  same 
vertical  line  with  it,  will  be  found  the  decimal  of  the  logarithm, 
to  which  prefix  its  proper  characteristic.  Thus,  log.  4790  z= 
a  0808792;  log.  -03745  =  2-5734518*. 

When  the  number  contains  more  than  four  figures,  find  tli* 
decimal  part  of  the  logarithm  of  the  first  four,  as  already  direct- 
ed ;  then  consider  the  remaining  figures  of  the  number  as  a  frac- 
tion, placing  a  decimal  point  before  them  ;  multiply  the  difference 
between  the  logarithm  already  found  and  the  next  greater  by  thia 
fraction  ;  and,  rejecting  as  many  figures  on  the  right  as  there  are 
dtscimals  in  the  multiplier,  add  the  product  to  the  decimal  of  ihe 
logarithm  corresponding  to  the  first  four  figures,  remembering  tc 
place  the  right  hand  figures  of  the  decimals  to  be  added  undei 
each  other ;  prefix  the  appropriate  characteristic,  and  the  result 
will  be  the  logarithm  sought. 

For  example,  in  finding  the  logarithm  of  3745I2G,  we  take  the 
decimal  part  of  the  logarithm  corresponding  to  3745,  and  add  to 
it  -120  of  the  difference  between  that  logarithm  and  the  neyt 
greater,  and  to  the  result  prefix  0  as  a  characteristic. 

The  reason  of  this  method  of  proceeding  will  be  seen  as  fel- 
lows. The  decimal  logarithm  of  3745000  is  -5734518;  t^ie 
next  greater  decimal  logarithm,*  corresponding  to  3740000,  is 
5735678.  The  difference  between  these  numbers  is  1000,  rnd 
the  difference  between  their  logarithms  is  '0001160;  moreover, 
3745126  exceeds  3745000  by  126.  Wherefore,  if,  when  the 
number  increases  1000,  the  logarithm  increases  -0001160,  r  hen 
the  number  increases  126,  the  logarithm  should  increase  .  /^g, 
-  -126,  of -0001160,  which  is  -0000146160;  retaining  onl'  sev 
HX  decimals,  we  have  '0000146,  which  added  to  '5734518   gives 

In  the  mor<»  e«cended  tables,  as  those  of  Callet,  four  figures  of  tl  nura- 
cer  are  found  in  the  left  hand  column,  and  the  fifth  at  the  top.  JVl  iover 
fropoftional  parts  of  the  differences,  are  placed  on  the  right  hand  sic      «^  the 


LVll.  USE    OF    LOGARITHMIC    TABLES.  267 

•5734G04;  to  this  sum  prefix  the  characteristic  G,  and  we  hare 
.eg.  3745126  =  6  5734664. 

The  result  would  evidently  have  been  the  same,  if  we  had  dis- 
regarded the  rank  of  the  decimals  in  the  difference  of  the  loga- 
rithms, nmltiplied  this  difference  by  '126,  rejected  the  three 
right  hand  figures  of  the  product,  and  added  the  reserved  part  of 
the  pioduct  to  65734518,  placing  the  right  hand  figure  updei 
the  8. 

In  like  manner,  we  shall  find  log.  327983  =:5'5158514  ;  also, 
log.   0379426  =  2  5791271. 

Remark.  This  mode  of  finding  the  logarithms  of  large  num- 
bers, as  well  as  that  to  be  given  for  finding  numbers  correspond- 
ing to  given  logarithms,  supposes  that  logarithms  increase  in  the 
same  ratio  as  the  numbers  themselves,  which,  though  not  strictly 
true,  is  nearer  the  truth,  the  greater  the  numbers  and  the  less 
their  difference  and  gives  results  sufficiently  accurate  for  most 
practical  purposes. 

Let  the  'earner  find  from  his  tables  the  logarithms  of  the  fol- 
lowing numbers 


1. 

1273. 

6. 

12710-63 

2. 

57293. 

7. 

274967. 

3. 

•01273. 

8. 

333  333 

4. 

•00279. 

9. 

435-501. 

5. 

327496. 

10. 

111-3734 

Art.  1G3.  To  find  the  number  corresponding  to  any  given 
logarithm. 

Look  for  the  decimal  part  of  the  logarithm  in  the  tab  e,  and 
if  it  be  found  exactly,  the  first  three  figures  of  the  number  will 
be  found  in  the  left  hand  column,  marked  N.,  in  the  same  hori- 
zontal line  with  the  logarithm,  and  the  fourth  at  the  top,  directly 
above  the  logarithm,  the  rank  of  the  figures  being  shown  by  the 
characteristic. 

Thus,  3-5860244  being  the  given  logarithm,  the  correspond 
mg  number  is  3855.     Had  the  characteristic  been  1,  the  nurn' 


268  USE  OF  LOGARITHMIC  TABLES.  lA  II 

ber  would  have  been  38'55 ;  had  it  been  6,  the  number  vvouIH 
have  been  3855000  and  had  it  been  2,  the  number  would  have 
been  03^55. 

If  the  decimal  part  of  the  logarithm  is  not  found  exactly  in  the 
table,  take  the  difference  between  the  given  logarithm  and  th<» 
next  less  tabular  logarithm,  for  a  numerator,  and  the  difference 
between  the  next  less  and  the  next  greater  tabular  logarithms,  for 
a  denominator.  Reduce  the  fraction  thus  formed  to  a  decimal, 
and,  rejecting  the  decimal  point,  place  the  result  at  the  right  of 
the  four  figures  corresponding  to  the  less  tabular  logarithm ; 
lastly,  place  a  decimal  point,  if  necessary,  according  to  the  char- 
acteristic of  the  given  logarithm. 

For  example,  let  2-4716423  be  the  logarithm,  the  correspond- 
ing number  to  which  we  wish  to  find.  The  next  less  decimal 
in  the  table  is  '4715851,  the  difference  between  which  and  the 
given  logarithm,  the  characteristic  being  neglected,  is  572  of 
the  lowest  order  of  decimals  in  the  logarithms ;  the  difference 
between  the  next  less  and  next  greater  tabular  logarithms,  is 
1466  of  the  lowest  order  of  decimals  in  the  logarithms  ;  reducing 
i^?¥F  ^^  ^  decimal,  we  have  the  figures  39,  which  placed  at  the 
right  of  2962,  the  figures  corresponding  to  •4715851,  gives 
296239 ;  but  as  the  characteristic  of  the  given  logarithm  is  2, 
we  point  off  three  figures  for  integers,  and  obtain  296239  for 
the  required  number. 

The  reason  of  this  method  is  obvious.  For,  if,  when  the  loga- 
rithm increases  1466,  the  number  increases  I  unit  of  any  order, 
when  the  logarithm  increases  572,  the  number  ought  to  increase 
i^-^wE  ^^  ^  ""'^  ^^  ^^®  same  order.  The  order  of  the  unit  of 
which  we  find  a  fractional  part,  is  always  determined  by  the 
characteristic.  In  the  example  just  given  we  found  the  fractional 
part  of  01,  viz.  "039 ;  but  had  the  characteristic  been  3,  the  frac- 
tion would  have  been  a  part  of  1. 

Let  the  learner  find  the  numbers  corresponding  to  the  follow- 
ing logarithms,  carrying  the  numbers  to  six  significant  figures 
when  the  decimals  are  not  found  exactly  in  the  tables. 


LVII.  USE  OF  LOGARITHMIC  TABLES.  269 

Common  tables  will    generally  give  numbers  with  sufficient 
accuracy  to  six  or  seven  figures. 


1. 

1-4771213. 

6 

02134445. 

2. 

3-3010300. 

7. 

i-4840150. 

3. 

1-4991217. 

8. 

27734667. 

4. 

3-1171167. 

9 

32276677. 

5. 

5-3458726. 

10. 

43334475. 

An.  164:.  Since  the  logarithm  of  a  vulgar  fraction  is  found 
by  subtracting  the  logarithm  of  the  denominator  from  that  of  the 
numerator,  it  follows  that  the  logarithm  of  any  proper  fraction, 
like  that  of  a  decimal,  must  be  negative.  But  we  ordinarily 
make  the  characteristic  only  negative. 

Thus,  log.  2^  =  log.  2  —  log.  257  =  0-3010300  —  2-4099331 

r=  389 10969.  In  order  that  we  may  be  able  to  subtract  the  lat- 
ter decimal  from  the  former,  we  may  suppose  the  characteristic 

0  of  the  logarithm  0-3010300  to  be  changed  into  —  1  +  1,  so 
that  the  decimal  -4099331  can  be  taken  from  the  positive  part  of 

1  +  1-3010300;  then  subtracting  the  2  from  —  1,  we  have  3 
for  a  characteristic.  Or,  as  is  more  commonly  done  in  subtrac- 
tion, after  having  borrowed  1  in  subtracting  the  left  hand  deci- 
mal, we  may  carry  I  to  the  2,  and  then  subtract  the  3,  which 

giv?  3,  the  same  as  before. 

Art.  100.  But  there  is  another  form  for  the  logarithm  of  any 
proper  fraction,  by  which  the  negative  characteristic  is  avoided. 
This  form  is  obtained  by  increasing  the  true  characteristic 
by  10. 

For  example,  the  logarithm  of  -3  is  1-4771213;  adding  10  to 
the  characteristic  and  reducing  gives  log.  -3=3  9-4771213.  In 
like  manner,  log.  -03  =  84771213 ;  and  log.   003  =  7-4771213. 

ilence,  in  this  way,  if  the  first  figure  of  the  decimal  is  tenths, 
the  characteristic  of  its  logarithm  is  9 ;  if  the  first  figure  is  hun« 
dredths,  the  characteristic  is  8  ;  if  the  first  figure  is  thousandths, 
the  characteristic  is  7,  and  so  on.     That  is : 
23* 


*'"U  USE    OF    LOGARITHMIC    TABLES.  LVH 

Tht  characteristic  of  the  logarithm  of  a  decimal  fraction,  is  IC 
di7niftished  by  as  many  units^  as  are  equal  to  the  number  of  places 
by  which  the  first  significant  figure  of  the  fraction  is  removea 
frmn  the  decimal  point.  « 

Likewise,  in  finding  the  logarithm  of  a  vulgar  fraction,  we 
may  increase  the  logarithm  of  the  numerator  by  10,  and  then 
subtract  the  logarithm  of  the  denominator.  Thus,  the  logarithm 
of  fIt  would,  in  this  way,  be  7-8910969. 

But  we  must  recollect,  that  every  such  logarithm  is,  in  fact,  10 
too  great,  and  that  the  result  of  any  operation  in  which  it  is  used, 
would  be  affected  accordingly. 

Art.  160.  In  division,  we  have  seen  that  the  logarithm  of 
the  divisor  is  to  be  subtracted  from  that  of  the  dividend  ;  but,  in- 
stead of  this,  we  may  add  the  arithmetical  complement  of  the  log- 
arithm of  the  divisor  to  the  logarithm  of  the  dividend,  dropping 
10  afterwards  in  the  result. 

The  arithmetical  complement  of  the  logarithm  of  a  number,  is 
what  remains,  after  the  logarithm  of  that  number  has  been  sub' 
traded  from  10 

Thus,  the  arithmetical  complement  of  the  logarithm  of  17,  is 
10  — log.  17=10— 12304489  =  8-7695511. 

The  arithmetical  complement  of  a  logarithm  may  be  found,  by 
biibtracting  the  first  right  hand  significant  figure  of  the  logarithm 
from  10,  and  all  the  others  from  9;  so  that  we  may,  if  we  please 
commence  the  subtraction  at  the  left  hand. 

Wc  must  bear  in  mind  that  each  arithmetical  complement 
added,  makes*  the  result  10  too  great,  and  allow  for  this  in  any 
operation,  in  which  arithmetical  complements  of  logarithms  are 
used. 

The  fact  that  adding  the  arithmetical  complement  of  a  loga- 
rithm and  afterwards  subtracting  10,  is  equivalent  to  subtracting 
the  logarithm  itself,  may  be  easily  proved. 

For,  let  /  represent  any  logarithm,  and  /'  a  logarithm  which  is 
to  be  subtracted  from  it;  the  result  would  be  / — /'.  Now 
the  ai-thmetical  complement  of  /'  is  10  —  /';  adding  this  to  /, 


LVII.  USE    OF    LOGARlTHMlJU    TABLES.  271 

we  have  ^+10  —  I' \  subtracting  10,  we  have /-|- 10  —  /'  — 10  , 
which  reduced  becomes  /  —  /',  the  same  result  as  when  I'  was 
subtracted  immediately  from  /. 

If  however  we  add  the  arithmetical  complement  of  the  loga- 
rithm of  a  fraction  with  its  characteristic  10  too  great,  the  result, 
without  dropping  10,  will  be  the  same  as  if  the  logarithm  of  the 
fraction  had  been  subtracted. 

To  prove  this,  let  /'  be  the  true  logarithm  of  any  fraction ; 
then  lO-f-Z'  would  be  its  logarithm  with  a  characteristic  10  too 
great ;  the  arithmetical  complement  of  this  is  10 —  10  —  /',  which 
added  to  any  logarithm  /,  gives  /+  10  —  10 — /'  or  /  —  /', 
which  is  precisely  the  same  as  if  /'  were  directly  subtracted 
from  /. 

Art.  1 67.  Let  the  learner  find  the  logarithms  of  the  follow- 
ing numbers.  When  the  fractions  are  less  than  unity,  the  loga- 
rithms may  be  given  in  both  forms,  viz  :  with  the  negative  char- 
acteristic, and  with  the  characteristic  10  too  great. 


1. 

•7234. 

6. 

¥/• 

o 

•00576. 

7. 

3A- 

3. 

•00087926. 

8. 

456+f. 

4 

h 

9. 

145J. 

5. 

Iff- 

10. 

•Nt^- 

Art.  168. 

Find  the  numbers 

correspor 

iding  to 

logarithms,  the  characteristics  being  each  10  too  great.  Six  sig- 
nificant figures  may  be  found  in  each  case,  when  the  decimal  j^it 
is  not  found  exactly  in  the  tables. 


L.     94371213. 

5.     5-4771213 

2.     87294179. 

6.     9-8878879. 

3.     6-3010300. 

7.     8-830037^ 

4.     26734217. 

8.     7-4378678. 

B.    It  will  be  observed 

that 

fmding  the  logarithm  of  a  \  ul- 

N. 

gar  fraction,  aid  then  obtainiiig  t;.5  number  corresponding  to 
th^t  logarithm,  converts  the  fract  3n  m*o  a  -^o^  ina' 


2'"2  APPLICATION    OF    LOiJARITHMS  LVlll 

SECTION     LVlll.. 

APPLICATION    OF    LOGARITHMS    TO    ARITHMETICAL    OPKRATIOMS 

Art.  109.    1.  Multiply  456  by  723. 

Log.  456  =  2  6589648 

Log.  723  —  2  8591383 

Prod.  z=z  329688 ---5-518I03L 

Adding  the  logarithms  of  456  and  723,  we  find  the  sum  to  be 
5  5181031 ;  we  then  find  from  the  tables  the  number  correspond- 
ing to  this  logarithm,  viz :  329688,  which  is  the  required  pro- 
duct. 

2.  Multiply  2678  by   03745. 

Log.    2678    =3-4278106 

Log.    03745  =  25734518 

Prod.  =  100  29-----  20012624. 

In  adding,  there  is  1  to  carry  when  we  arrive  at  the  character- 
istics; this  1  is  positive,  and,  being  added  along  with  the  3  and 
8,  gives  for  a  characteristic  4  —  2  or  2. 

The  same  without  the  negative  characteristic. 

Log.    2678    ==  3-4278106 

Log.    03745  =  8-5734518 

12  0012624 

Subtract--- 10 


Prod.  =  100-29 2  0012624. 

It  would    have   saved   labor  to  drop  the  10  at   the  time  of 
adding. 

3.  Multiply  -0374  by  277. 

Log.  -0374  =  2-5728716 
Log.  -277    =  1-4424798 
Prod.  =   0103598 ---2-0153514. 
The  figures  answering  to  the  decimal  of  the  resulting  loga^ 
nthm  are  103598;  but  since  the  characteristic  is  2,  the  first 


LVIII.  TO    ARITHMETICAL    OPERATIONS.  273 

figure  of  the  number  must  be  hundredths,  therefore  a  zero  pre- 
ceded by  the  decimal  point,  must  be  placed  before  the  figures 
found. 

The  same  without  negative  characteristics. 

Log.  •0374  =  8-5728716 
Log.  -277    =  9-4424798 
Prod.  =  •0103598---8-0153514. 
In  the  suiu  of  the  logarithms,  the  characteristic  becomes,  in 
fact,  two  tens  too  great ;  but  we  drop  only  one  of  them,  and  then 
the    characteristic  8  shows  that  the  first  figure  of  the  number 
must  be  hundredths. . 

4.  Divide  48945  by  65. 


5  By 


Log.  48945  =  4  6897083 
Log.  65        z=  18129134  S  By  subtraction, 
auot  =753 2-8767949. 

The  same  with  the  arithmetical  complement  of  the  logarithm 
of  the  divisor.  The  contracted  expression,  comp.  log.^  will  some- 
times be  used  to  signify  arithmetical  complement  of  the  loga- 
rithm. 

Log.      48945  =  4-6897083  ^ 
Comp.  log.  65  =  8-1870866  5  Add. 

auot.  z=  753 2-8767949 

5.  Divide  775  by  -025. 

Log.  775  =  2-8893017 

Log.   025  =  ^-3979400 

auot.  =  31000  -  -  -  4  4913617. 

The  sign  of  2,  in  subtracting  is  changed  to  -|-,  and  then  the 
two  characteristics  are  added. 

The  same  without  the  negative  characteristic,  and  with  the 
comp.  log.  of  the  divisor. 

Log.   775  = 2-8893017 

Log.  -025  =  8-3979400,  comp.  log.  =  1-6020600 
auot.  =  31000 4  4913617. 


274  APPLICATION    OF    LOGARITHMS  LVIll 

In  this  example,  the  resulting  characteristic  is  not  too  great 
because  the  logarithm  of  the  fraction  was  taken  without  the  neg 
ative  characteristic. 

6.  Divide  005127  by  0559. 

Log.  •005127  =  3-7098633 

Log.    0559     ^z=z  2-7474118 


Quot.  =  -091717 2  9624515. 

In  subtracting,  there  is  1  to  carry  to  the  2,  which  makes  it  I, 
and  this  subtracted  from  3,  gives  3  +  1  or  2  for  a  characteristic 
Or,  if  the  learner  is  accustomed,  when  he  borrows  1,  to  dimin- 
ish by  1  the  next  figure  in  the  minuend,  he  will  take  1  from  3, 
which  gives  4;  from  this  he  will  then  subtract  2,  and  obtain 
4  +  2,  or  2. 

The  same  with  positive  characteristics  and  comp.  log.  of  the 
divisor. 

Log.  -005127  =  -    -    - 7-7098633 

Log.  -0559  =  8-7474118,  comp.  log.  =  1-2525882 
auot.  =  -091717 8-9624515 

In  this  case  the  resulting  logarithm  is  10  too  great,  because 
the  logarithm  of  the  dividend  was  taken  with  a  positive  chaiao 
teristic. 

7.  Required  the  third  power  of  27. 

Log.  27=  I  4313638 

3     Mult,  by  3. 

Power  =  19683  -  -  4-2940914. 

8.  Required  the  third  ppwer  of  -271. 

Log.  -271  =  1-4329693 
3 

Power  =  -0199025 1 1  -  -  -  -  2-2989079. 


LVIIl  TO    ARITHMETICAL    OPERATIONS.  275 

The  same  with  positive  characteristics. 
Log.  -271  =  9-4329693 


Power  =  -019902511  -  -82989079. 
By  thi«   last  method,  the  characteristic  after  multiplication, 
becomes  28,  which  is  three  lOs  or  30  too  grtat ;  dropping  two 
10s  or  20,  we  have  the  characteristic  8,  which  shows  that  the 
fiist  figure  of  the  number  is  hui  dredths. 
9    Required  the  fifth  root  of  15. 

Log.  15  =  1  1760913(5     Divide  by  5 
Rpot  =:  1-71877  -  -  -  0  2352183. 

10.  Find  the  third  root  of  -000729. 

Log.  -000729  =  4-8627275  =  6  +  2-8627275(3 
Root  =09 2-9542425. 

In  this  question  a  difficulty  occurs  in  dividing  the  logarithm 

48627275,  since  the  integral  and  fractional  parts  have  different 
signs,  and  the  negative  characteristic  is  not  divisible  by  3.      To 

obviate  this  difficulty,  add  2-)- 2,  which  is  zero,  to  the  charac- 
teristic; the  logarithm  then  becomes  6  +  28627275.  Dividing 
now  the  negative  and  positive  parts  separately,  we  have  the  re- 
sult as  above. 

In  all  cases  of  finding  the  root  of  a  fraction,  if  its  logarithm  ia 
taken  with  a  negative  exponent,  and  that  exponent  is  not  divisi- 
ble by  the  number  expressing  the  degree  of  the  root,  we  must 

make  it  so,  by  guding  to  the  logarithm  1  +  1,  2  +2,  3  -f-  3,  or 
*ome  equivalent  expression. 

T'lifi  same  with  positive  characteristics. 

Log.    000729  z=z  6-8627275 

20  -    -    -    -  Add. 
26-8627275(3 
Root  =   Oi  -8  9542425. 


276  APPLICATION    OF    LOGARITHMS  I  Vlll 

By  the  second  method,  the  logarithm  when  first  found,  is  too 
great  by  10;  we  then  add  two  more  10s,  which  makes  it  three 
lOs  too  great ;  this  divided  by  3  gives  a  result  10  too  great  as  re- 
quired. 

Whenever  we  use  the  positive  characteristic  in  finding  the  root 
of  a  fraction,  before  dividing  the  logarithm,  it  is  necessary  to 
make  the  characteristic  as  many  10s  too  great  as  there  are  units 
in  the  number  which  marks  the  degree  of  the  root.  The  divi- 
81  jn  will  then  leave  the  result  one  10  too  great. 

11.  Find  the  value  of  x  in  the  expression,  x=z  (^)^. 

Log.  2  = 0-3010300  • 

Comp.  log.  7  r=     -     - 9- 1549020 

Log.  f  -  -  char.  1 0  too  great,  :^  9-4559320 

3 

Log.  (f )3 -  -  char,  three  10s  too  great,  =  283677960 

20 Add. 

48-3677"960(5    * 
X  =  -471584    - 9  6735592. 

12.     Find   the   value   of  x   in  the   expression,   x  = 
/45  .  13  .  (•75)\f 
Vl9.  117.  11  /  • 

Log.  45  = 1-6532125 

Log.   13  = 1-1139434 

Log.  -75  = 9-8750613 

Comp.  log.    19  =  8  7212464 

Comp.  log    117  =  7-9318141 

Comp.  log.    1 1   =  8;9586073 

8^2538850 

3 

24-7616550 

10 

34-7616550(4 
%  z=  -0490245  -    -  -  6^904137. 


LVIIL 


TO    ARITHMETICAl.    JPERATIONS. 


277 


In  this  question  we  have  used  the  logarithm  of  one  fract'cn, 
with  the  increased  characteristic,  and  three  corap.  logS.  of  whole 
numbers ;  the  sum  of  the  six  logarithms  added,  will  therefore  be 
40  too  great.  Dropping  30,  multiplying  by  3,  adding  10  to  the 
product,  and  dividing  this  sum  by  4,  will  leave  the  final  'oga- 
rithm  10  too  great. 

13     Find  the  value  of  x  in  the  expression. 

(12)=.(|)*.V'i 

Log.  2  =  0-30t0300 

Comp.  log.  3  =  9-5228787 


9-8239087 

10- 

19-8239087  (2 

Log.v/S 

—  Q'Ql  lQ'^/1*^   .   .  .  .  . 

UQ1IU<!LL'!| 

Log.  7 

=  0-8450980 

Comp.  log. 

8  =  90969100 
9-9420080 
20- 
29-9420080  (3 

Log.^i 

=  9-9806693 

. . . 

-  .  g-9806693 

Log.  -075 
Log.  12 

-    .  R'ftT'^nfii'i 

=  10791812 
3 

Log.  (12)3 

=  3  2375436-.  comp. 

log. 

=  6-7624564 

Log.  3 

=:=  0-4771213 

Comp.  log. 

5  =  9-3010300 

Log.l 

=  9-7781513 
4 

W-  (lY 

—  9  1126052 -- comp. 

log. 

=:  0-8873948 

24 

' 

278  APPLICATION    OF    LO<iARITHMS  LVIIT 

Lo^  —  0-6020600 

Comp.  lov  9  —  904575-5'^ 
Log.  I  ^  9-647817S 

40- 

49^78175  (5 

Log.  ^1"        =  9-9295635  -  -  -  comp.  log.     :=  0  0704365 

We  now  add  the  several  results  which  are  carried  out  to  thd 
fight 

W-  v/f  =  ^'^^  ^^^"^^ 

Log.  ^1  =9-98D6693 

Log.  -075  =  8-8750613 

Comp.  log.  (12)3  =z  6-7624564 
Comp.  log.  (1)4    =  0-8873948 

Comp.  log.  ^1     =  00704365 
X  =  00030695  6-4870726. 

Some  labor  might  have  been  saved  in  this  problem,  by  substi 
tuting  equivalents  for  several  of  the  quantitie*  viz :  '875  for  f , 
1728  for  (12)3,  and  -6  for  f.  But  the  object  was,  to  exhibit  the 
general  mode  of  proceeding,  and  not  the  shortest  for  this  partic- 
ular case. 

Although  in  several  of  the  preceding  problems,  logarithms  of 
fractions  have  been  used  in  both  forms,  it  is  advisable,  in  most 
cases,  to  use  the  increased  characteristic ;  especially  as  the  learner 
who  is  to  study  Trigonometry,  will  have  occasion  to  use  tables  in 
vhich  every  characteristic  is  10  too  great. 

Perform  the  following  questions  by  means  of  logarithms. 

14.  Multiply  37-153  by  4086. 

15.  Multiply  257-3  by  300. 

16.  Multiply  567  by  -572. 

17.  Multiply  -0387  bv  093. 

18.  Multiply  f  bv  11-5756, 

19.  Multiply  f^l  bv  9| 


.^VIIJ  TO    ARITHMETICAL    OPERATIONS.  27ll» 

20  Multiply  147f  by  24^. 

21.  Find  the  product  of  375,  325,  and  03756 

22.  Divide     12783  by  256. 

23.  Divide  147324  by  248333. 

24.  Divide    22563  by        0473. 

25.  Divide       0743  by       -3967. 

26.  Divide  ^  by  yf^. 

27.  Divide       126f  by  17f. 

28.  Find  the  4th  power  of  2*73. 

29.  Find  the  3d  power  of  916. 

30.  Find  the  5th  power  of  -03. 

31.  Find  the  5th  power  of  2J. 

32.  Find  the  2d  root  of  5.     • 

33.  Find  the  3d  root  of  423. 

34.  Find  the  3d  root  of  -0756 

35.  Find  the  4th  root  of  -37 

36.  Find  the  7th  root  of  '951 

37.  Find  the  5th  root  of  f  |. 

38.  Find  the  value  of  (f  )^ 

39.  Find  the  value  of  (||f  )^. 


40.  Find  the  value  of  ^49  .  f  .  (-0673). 

41.  Find  the  value  of  (f  |)*  .  (iH)^- 

42.  Find  the  value  of  ^^  .  \/J^. 
^3  .  (073) .  256 


43.     Find  the  value  of 


V^f  •  (A)^  .  (-056)^ 


44.     Find  the  value  of  x  in  the  equation,  55*=  493 

Sucn  an  expression  as  55^,  in  which  the  exponent  is  unknown, 
is  called  an  exponential  quantity. 

Since  the  logarithm  of  any  power  of  a  quantity,  is  found  by 
multiplying  the  logarithm  of  that  quantity  by  the  number  whicVi 
expresses  the  degree  of  the  power,  we  have,  in  the  present  case, 
by  taking  the  logarithms  of  both  members. 


280  APPLICATION    OF    LOUARIT51MS  LVIM 

X  X  log.  55  =  log.  493,  or, 

X  X  1-7403627  zr:  2-6928469.     Hence, 

The   division,    performed    in    i comraon  way,  gives   x  = 

.5473. 

But  we  may  take  the  logarithms  of  these  logarithms,  as  we 
would  of  any  other  numbers,  and  perform  the  division  as  usuai 
with  logarithms. 

Log.  2-6928469  =  04302 1 17 
Comp.  log.  1  -7403627  =  97593603 
X  —  1  5473  -  - 01895720. 

Let  the  learner  perfoim  the  following  questions,  finding  fi?6 
l^gures  in  the  answer  to  each. 

45.  Find  2  in  4^  =  27. 

46.  Find  a;  in  7^  =  9. 

47.  Find  x  in  12 '  =  44. 

In  the  last  question,  raise  both  members  to  tne  ztn  power, 
<^hich  gives  44*  =  12^,  or  44*  r=:  1728 ;  the  value  of  x  may  then 
DC  found  as  in  th3  preceding  examples. 

48.    Find  x  in  the  proportion,  720  :  196  =  155*5  :  x. 

We    know    from   the    principles    of   proportion   that    x  = 

'-— ;  hence,  we  are  to  add  together  the  logarithms  of  the 

means,  and  the  comp.  log  of  the  first  term.     We  may  therefore 
begin  with  the  first  term. 

Comp.  log.  720  -  7- 1426675 
.  Log.  196  —  3  2922561 

Log.  155-5         =  2-1917304 

X  =  42  33 r6266540. 

For  the  coi,venience  of  applying  logarithuio,  the  terms  of  a 
proportion  may  be  placed  under  each  other,  caie  being  taken  to 
change  the  order  of  the  terms,  if  necessary,  so  that  the  unknown 
shall  stand  last,  and  to  use  the  comp.  log.  of  the  first  term  in  that 
arrangeiTiPn*. 


LVIIl.  TO    ARITHMETICAL    OPERATIONS  881 

49.  Find  x  in  the  proportion,  15  :  x  =z  100  :  47. 

50.  Find  a  mean  proportional  between  12'5  and  75*o3. 
5i.    Insert  four  mean  proportionals  between?  and  20. 

52.    Required  the  sum  of  a  progression  by  quotient,  the  firs 
4jrm  being  5,  the  ratio  4,  and  the  number  of  terms  6. 

Substituting   the   given  numbers   in   the   formula, 

(7"— 1)           ,          c,      5(46  —  1) 
-^^ — -,  we  have  S=z — ^— ^r ^. 

q  —  1  o 

Log.  4  =  0-6020600 

6 

46  =  4096-  -  -. 3-6123600 

1 

46  —  1  =  4095,  its  log.  =  3-6122539  ' 
Log.  5             =  0-6989700 
Comp.  log.  3  =  9-5228787 
5?=  6825 3-8341026. 

Art.  170.    We  may  now  solve  the  four  questions  in  progres- 
sion by  quotient  mentioned  in  Art.  151,  assuming  the  formulae 

/ 1=  a  o  *"  1,  and  S  =z — .     The  solution  of  one  of  them  will 

q  —  l 

-e  given,  and  that  of  the  others  will  be  left  as  an  exercise  to  the 

earner. 

1.    Given  a,  q  and  /;  to  find  S  and  n. 

/       /J 

The  value  of  S  is  already  given,  viz  :  iS  =:^ ~. 

To  find  n\  the  equation,  /=  a q^~^,  gives,  in  succession^ 

,-.=1. 

{n  —  1 )  log.  q  =  log.  f  —  j  =:  log.  / —  log.  a ;  hence, 

log.  / — loff.  a 

n  —  1  =  -^-j — ,  and 

log.  q 

log.  / — log.  a 


log.  q 


24* 


282  COMPOUND    INTEREST*  UX 

5     Gi  en  a,  /  and  S;  find  q  and  «. 
S     Gi  en  a,  y  and  S ;  find  /  and  n. 
4    Given  §-,  /  and  S;  find  a  and  n. 
The  follow  ng  questions  may  be  solved  by  means  of  Jie  for 
mulae  obtain©  i  from  the  four  preceding  problems. 

5.  The  first  term  of  a  progression  by  quotient  being  3,  the 
ratio  2.  and  the  last  term  6144 ;  required  the  sum  and  the  num» 
ber  of  terms. 

6.  The  first  term  of  a  progression  by  quotient  is  6,  the  last 
term  13122,  and  the  sum  19680;  required  the  ratio  and  the  num- 
ber of  terms. 

7.  The  first  term  of  a  progression  by  quotient  being  9,  the 
ratio  3,  and  the  sum  265716;  required  the  number  of  terms  and 
the  last  term. 

8.  The  sui  \  of  a  progression  by  quotient  being  6560,  the  ratio 
3,  and  the  la^*  term  4374;  required  the  first  term  and  the  num 
ber  of  terms. 


SECTION    LIX. 


COMPOUND    INTERS8T. 


Art.  171.  Let  p  represent  any  sum  of  money  put  at  com- 
pound interest,  for  a  number  t  of  years,  at  the  annual  rate  of  r 
per  cent.,  r  being  a  decimal,  as  05  or  *06.  It  is  required  to 
find  the  amount,  which  we  represent  by  A. 

It  is  manifest,  that,  if  any  principal  be  multiplied  by  1  -|-  the 
rate,  the  product  will  be  the  amount  for  one  year ;  for  this  is  the 
same  as  mu.t.plying  the  principal  by  the  rate,  which  gives  the 
interest  for  one  year,  and  adding  the  result  to  the  principal. 
Thus,  the  a  ount  of  $10,  for  a  year  at.6  per  tent.,  is  10(1*06) 
or  $10-60. 

The  amount,  then,  of^  dollars  for  one  year,  is  p(l  -f-'*) ;  this 
«  the  capital  for  the  second  year,  and,  to  obtain  the  amount  at 


LIX.  COMPOUND     INTEREST.  28* 

the  end  of  tftat  year,  we  mast  multiply  this  capital  by  1  -|- «• 
which  gives  jp  (1  +^)^;  this  being  the  capital  for  the  third  year 
and  being  multiplied  by  1  +  r,  gives,  for  the  amount  at  the  en6 
of  the  third  year,  p  (I  -\-r)^.  In  like  manner,  the  amount  at  the 
end  of  the  fourth  year  is  p{l  -{-ry;  tnat  at  the  end  of  the  fifth 
year  is  p  ( 1  -f-  '*)^« 

The  amount  in  any  case,  therefore,  is  found  by  raising  1  -|- 
the  rate  tr»  the  power  denoted  by  the  number  of  years,  and  mul 
tiplying  tfte  result  by  the  principal. 

The  formula  for  the  amount,  therefore,  is 

A=p{l  \-ry. 

1.  Required  the  amount  of  $/J0,  for  4  years  at  6  per  cent 
Bompound  interest. 

In  this  question,  p  =z  750,  r  =  '06,  and  t  =z4.  Substituting 
hese  numbers  in  the  formula,  we  have  A  =z  750  (106)'*. 

Log.  106      =  00253059 

4 

Log.  (1-06)4  =  01012236 

Log.  750       z=  2-8750613 

A  =  $946-858 29762849. 

2.  Required  the  amount  of  $1050,  for  5^  years  at  5  per  cent, 
compound  interest. 

Log.  105        =  0-0211893 

.      ^ 

0-1059465 

0105946 

Log.  (105)^*  =  oTiBHTn 

Log.  1050        =  30211893 

A  =  $1373- 19 31377304. 

It  is  common  with  merchants,  to  find  the  amount  for  the  num- 
ber of  whole  years,  and  then  find,  at  simple  interest,  the  amount 
of  that  s&m  for  the  fractional  part  of  a  year.  According  to  this 
method,  the  process  by  logiirithins  would  be  as  follows. 


284  COMPOUND    INTEREST.  LIX 

Log.  1-05  =    «0211893 

5 

Log.  (105)5                    _  01059465     . 
Log.  1050                       z.  30211893 
Log.  of  am.  for  5  years  r"-  31271358 
Log.  (1025)                    ^  00107239 
^  =  $1373  598 31378597. 

After  having  found  the  logarithm  of  the  amount  for  5  years, 
we  add  to  it  the  logarithm  of  1*025^  that  is,  of  1  +  the  rate  for 
six  months. 

The  last  result  exceeds  that  obtained  in  the  previous  solution 
by  $0408.  In  succeeding  questions  the  former  method  may  be 
pursued. 

Any  three  of  the  four  quantities  in  the  equation,  A  =^  (l-|-r)  , 
being  known,  the  remaining  one  may  be  found.  Making  p,  r 
and  t  successively  the  unknown  quantity,  we  obtain  the  follow 
ing  formulae. 

A 
P 


{l  +  rY 

-if)--' 


'-log.  (l+r)- 
3.  What  sum  must  be  put  at  interest,  the  rate  being  G  pel 
cent.,  in  order  to  amount  to  $1287  in  4  years? 

In  this  question  p  is  to   be    found,  and  the  formula,  p  = 

T— -j- — -f  by  the  substitution  of  the  given  quantities,  becomes 

1287  ,      ^ 

Log.  106      =  00253059 

4 

Log.  (106)4  =  01012236-  -  -comp.  log.  =  98987764 

Log.  1287     = 31095785 

^  =  $1019-424 3  0083545* 


LIX-  COMPOUND    INTEREST.  28C 

The  value  *)f  p  iii  this  example  is  called  the  present  worth 

4    "^he  principal  $400  amounts,  in  9  years,  at  *:ompound  in- 
•;e~es%  .o  $569*333 ;  required  the  rate  per  cent. 

Subs*i'.uting  in  the  formula,  r  =  (  —  1  <  —  1,    we  have  «•  = 
/56.)-33^>^^ 

i~4oo"7 

Log.  569-333     =  27553664 
Comp.  log.  400  =  7-3979400 

01^533064(9 
14-r  =  104-    -    -    -  0-0170340." 

1 

r=     04. 

6.  ilnv  many  years  must  $1000  remain  at  compound  intere^  , 
tie  rate  ueing  6  per  cent.,  in  order  to  amount  to  $1191'016T 

■««-(f)        ^ 

Substitutmg  ^^  -he  formula,  t  =: . -r-^, — .-,  we  obtain  ^  — 

'   log.  (1-06)     • 

Log.     1 191016  —  30759176 
Comp.  log.  1000  =  7-0000000 


-^t;) 


00759176 


Log.  lOi'.  —  00253059. 

Hence,  t  =  :§MHf  =  ^  years. 

Or,  performing  this  layt  division  by  logarithms   we  have 
Log.   0759176  =  8-8803424 

•      Comp.  log.  -0253059  =:  1-596778^ 

/  =  3 0-4771207. 

,o  this  question  the  operation  would  have  been  shorter,  if  we 
Had  divided  1191  016  by  1000,  before  applying  logarithms 


i86  COMPOUND  INTEREST.  LIX 

6.  Find  the  amount,  at  compound  interest^  of  $357'50,  for  8 
years  at  6  per  cent. 

7.  Find  the  amount  of  $1573  for  4  years,  at  5^  per  cent,  com 
pound  interest. 

8.  Required  the  compound  interest  on  $1000,  for  7  years  and 
4  months  at  4  per  cent. 

9.  What  sum  of  money  will,  in  6  years,  at  "  per  cent,  com 
pound  interest,  amount  to  $*2745-90? 

10.  What  sum  of  money  will  amount,  in  10  years,  to  $417  712, 
O'^mpound  interest  being  reckoned  at  6  per  cent.  ? 

11.  In  how  many  years  will  $75  amount  to  $149*495,  atSfer 
cent,  compound  interest? 

12.  A  principal  of  $108'50  amounted,  in  12  years,  at  com- 
pound interest,  to  $22045;  what  was  the  rate  per  cent.? 

13.  In  what  time  would  any  sum  be  doubled  at  compound  in- 
terest, the  rate  being  6  per  cent.  ? 

In  this  question  the  amount  is  to  become  double  the  princi- 
pal; therefore,  in  the  formula  for  ty  we  substitute  2p  instead 

of  Ay  which   ffives   t  =  z —-^. — r,  or,  by  reduction,  t  = 

^  log.  (l+r/  ^ 

log.S 

log.  ( i  f-  rV 

14.  In  how  many  years  will  any  sum,  at  compound  interest 
be  tripled,  the  rate  being  6  per  cent.  ? 

15  I'l  how  many  years  will  any  sum  be  doubled,  at  5  pei 
cent,  compound  interest? 

16.  What  would  $357  amount  to  in  10  years  at  compound  in- 
terest, the  interest  b6ing  reckoned  semi-annually,  at  the  rate  of 
6  per  cent,  a  year  ? 

17.  The  population  of  Boston  in  1830  was  61*^92;  what  was 
it  in  1840,  supposing  the  annual  rate  of  inert  J.»e  ja  >.e  3^J>r  ner 
rent.  ? 

18.  The  population  of  P\.ladeipnia  m  1830  was  lfeS797,  and 
»>  1840  it  was  258832:  what  was  the  annual  rate  of  increase  ? 


LIX.  COMPOUND    INTEREST.  287 

19.  In  1830  New  York  contained  202589,  and  in  184C  t  «onf 
ained  312234  inhabitants;  if  the  population  continue  to  increase 
at  the  same  rate  as  it  did  from  1830  to  1840,  in  how  many  years 
from  the  latter  date  will  it  amount  to  lOOOOOO  r 

Art.  179.  1.  A  man  saves  annually  $300  which,  at  the  end 
of  each  year,  he  deposites  in  a  bank,  and  is  allowed  5  per  cent, 
compound  interest.  How  much  would  be  due  him  froin  the 
bank,  at  the  end  of  12  years  from  the  time  of  the  first  deposit? 

To  generalize  this  question,  let  a  be  the  sum  annually  depos- 
ited, t  the  time,  and  r  the  rate.  Then  the  amount  of  the  sum 
first  deposited  would,  according  to  the  principles  already  given, 
be  a  (1 -(-'•)*•  The  second  deposit  remaining  in  the  bank  one 
year  less,  would  amount  to  a(l-|-r)*~i.  The  amount  of  the 
hird  deposit  would  be  a(l-j-r)*~2,  and  so  on.  The  last  de- 
posit but  one,  remaining  in  the  bank  two  years,  would  amount 
to  a  (1  -(- r)'2 ;  and  the  last  deposit  would  amount  to  a(l  +r). 

B  °nce,  if  A  represent  the  gross  amount,  we  have 

A=«(l+r)  +  «(l+r)2+ +«(l+ry-2^ 

^(l+r)'-i  +  a(l+ry. 

The  second  member  of  this  equation  is  a  progression  by  quo- 
lent,  in  which  the  first  term  is  a(l  +  r),  the  ratio  1  +  r,  and 

ie  last  terra  a  (1  +  r)*.  In  the  formula,  >S^=  — — r-,  substi- 
tuting A  instead  of  >S^,  a(\  +  r)*  instead  of  /,  a{\-\-r)  instea« 
of  a,  and  \-\-r  instead  of  9,  we  have 

^ 1  +  r-. •  °' 

_4^«(l+'-)[(l-t-'-)'-l]. 
r 

Substitutmg  in  this  formula  the  numbers  given  in  the  question 

Droposed,  we  have 

_  300,(105)  [(l'05)ia_i] 

05 


288  ANNUITIES.  LX 

In  applying  logarithms,  it  is  best  to  commence  with  the  quau 
•  ity  between  the  brackets. 

Log.  1()5  =  0021 1893 

12 

( 1  05)^2  _.  1  -795856  -  -  0-2542716 
1 


Log.     -795856  =  99008345 
Log.  1-05  =  00211893 

Log.  300  ==:  2-4771213 

Comp.  log.    05  =  1-3010300 
A  =1  $5013-893  ....  3*7001751. 

2.  If  a  young  man,  by  omitting  some  useless  expense,  saves  25 
•?ents  every  day,  and,  at  the  end  of  each  year,  deposits'his  savings 
m  an  institution  which  allows  6  per  cent,  compound  interest,  how 
»iiuch  would  be  due  him  from  the  institution,  at  the  end  of  20 
years  from  the  time  of  the  first  deposit,  a  year  being  considered 
365  days] 


SECTION    LX. 


ANNUITIES. 


Art.  173.  An  annuity  is  a  certain  sum  of  money  payable  an- 
lually,  or  at  other  regular  periods,  for  a  stated  number  of  years, 
or  during  a  person's  life,  or  forever.  The  following  question  is 
one  of  annuities. 

1.  A  man  wishes  to  put  at  compound  interest  such  a  sum  of 
money,  as  will  afford  him  annually  $500  for  20  years,  at  the  end 
of  which  time  the  principal  and  interest  shall  be  exhausted 
What  sum  must  he  put  at  interest,  the  rate  being  6  per  cent,  v 

It  is  manifest  that  the  amount  of  all  he  rpr;eives,  must  ue  thf 
same  %8  the  amount  of  the  sum  put  at  interest. 


LX.  ANNUITIES.  289 

To  generalize  this  question,  let  a  be  the  sum  received  annu' 
ally,  r  the  rate  of  interest,  and  t  the  time. 

As  the  first  sum  is  drawn  out  at  the  end  of  the  first  year,  the 
drawer  must  be  considered  as  having  received,  at  the  expiration 
of  the  whole  time,  the  amount  of  that  sum  at  compound  interest 
for  t — I  years,  which  according  to  Art.  171,  is  a(l +'*)'""• 
In  like  manner,  that  drawn  out  at  the  end  of  the  second  year, 
amounts  to  a(l  -\-ry-~^;  that  at  the  end  of  the  third  year,  to 
a  (1  -\-ry~^y  and  so  on;  the  sum  drawn  at  the  end  of  the  last 
jear  is  simply  a. 

The  gi  OSS  amount  of  the  whole  drawn  out,  is,  therefore, 

«(l  +  ry-i_|-a(|-|-r)'-2  +  a(l  +  r)*-3+  .  .  .  .+ 
a{\-^rY  ^  a(i-\-r)  -\-  a;  or,  by  a  change  in  the  order  of 
arrangement, 

a  +  a{\  +  r)  +  a{\+Yf+  .  .  ,  .  +«(l  +  ry-3  + 
aa+r)«-2  +  a(l+ry-i. 

This  is  a  progression  by  quotient,  in  which  the  first  term  is  a, 
the  ratio  1  -\-r,  and  the  last  term  a(l+r)*-^     Substituting 

these  in  the  formula,  <Sf=r — ,  we  have 

^^a(\+rY-y(i  +  r)~a  ^  a(\J^ry-a 

«[(I+rr-l] 

Now  let  A  be  the  sum  put  at  interest.  This  would  amount 
in  t  years  to  ^  (1  -|-  r)^ ;  and  since  this  amount  must  be  equal  to 
that  of  the  several  sums  drawn  out,  we  have 

^(I+.y  =  "[('+'->'-'];  hence, 

Substituting  the  numbers  given  in  the  proposed  ouestion.  v^« 

have  4_500[('-06r-l]  . 
have.l-        ^_-^.-_ -_  . 

25 


290  ANNUITIES.  LX 

Log.  106        =  00253059 

20 

(106)20  =  3-20714  -  -  0-5061180 

1 

hog.  2.20714  =  0-3438299 

Log.  500         =  2-6989700 

Comp.  log.  -06         =  1-2218487 

Comp.  log.  ( 1  -06)2*^  =  9-4938820 

A  =:  $5734-963 37585306 

In  the  equation,  A  =         ,7"  .     w — K  we  may  make  eithei 
r(l-f-r)*  •' 

of  the  quantities,  A^  a,  t  and  r,  the  unknown.     Thus,  to  find  o 

we  have  successively,  a  [(1  +  r)'  —  1]  =  Ar  (I  -\-ry ; 

Arji+r)' 

"--(T+77:rr-  • 

To  find  t  we  obtain    successively  fi-om  the   equation,  A  =■ 
a[n  +  r)'-l] 
r(l  +  r)'      • 

a(l+r)'  —  a  =  Ar(l  +  r)'; 
a{l  +r)'  —  Ar  (1  +r)'=  a; 
{a  —  Ar)(l+rY  =  a; 

<Xlog.(l  +  r)  =  log.(^-j-Ji 

'-     log.  (I  +  r)    • 
To  find  r  would  be  too  difficult  for  the  design  of  this  treatise 

2,  If  a  person  deposite  $5000  in  an  annuity  office,  how  much 
can  he  draw  annually,  if  the  annuity  is  to  continue  10  years, 
compound  interest  beimg  reckoned  at  5  per  cent  ? 


LX.  ANNUITIES.  291 

In  this  question,  a  is  the  unknown  quantity,  and  the  formula 
for  a,  by  substitution,  gives 

5000.  (-05)  (1-05)10  ^  250(105)10 
^""        (1-05)10  — 1         —  (l'05)iO  — 1* 

In  applying  logarithms,  it  is  best,  in  this  case,  to  commence 
with  the  denominator. 

Log.  1-05        =  00211893 

10 

(105)10  _  162889  -  -  0-2118930 

1 

Comp.  log.    -62889  =  0-2014253 

Log.  250       =  2-3979400 

Log.  (1-05)10  =  0  2118930 

a  =  $647-527 2*8112583. 

3.  A  man  deposits  in  an  aii»./ity  office  $7500,  for  which  5^ 
per  cent,  compound  interest  is  allowed ;  in  how  many  years  will 
it  be  exhausted,  if  he  draws  out  annually  $750? 

The  formula  for  ^,  by  substitution,  gives  t  = 

I        (  ^^^  \  1       ( J^\ 

^^*  V750  —  (  055).  7500/        u       j      •       .       ^^A337-50/ 

-, /,  ^g  ' ,  or,  by  reduction,  t  =  ~. ,^  ^--.  . 

log.  (1-055)  ^     y    J  log  (1-055) 

Log.  750  =  2-8750613 

Comp.  log.  337-50  =  7  4717262 

Log.  (t^:^)  =0^467875 

Log.  1055  =  0-0232525.     Hence, 

'  -—  .^2  3  2  5  2&' 

Log.  -3467875  =:  9-5400634 
Comp.  log.    0232525  =  1 -6335303 

t  =  14-914  years 11735937 ; 

or,  t  =z  14:  years,  10  months,  and  29  days. 

4.  A  gentleman  wishes  to  purchase  an  annuity,  which  shall 
afford  him  $500  annually  for  30  years ;  how  much  must  he  pay 
if  he  is  allowed  5  per  cent,  interest  ? 


292  MISCELLANEOUS    QUESTIONb. 

5.  How  much  must  be  given  for  an  annuity  to  last  <J0  years, 
if  S300  are  to  be  drawn  semi-annually,  and  interest  be  allowed 
at  the'  rate  of  4^  per  cent,  a  year  1 

6.  A  gentleman  purchases  an  annuity  for  the  benefit  of  his 
family  after  his  decease,  and  pays  $10000.  Three  years  from 
the  date  of  the  purchase  he  dies,  and  then  the  annuity  comes 
into  operation.  How  much  must  the  family  draw  out  annually, 
so  as  to  exhaust  the  annuity  in  15  years  from  the  time  it  com^ 
mences,  if  54  per  cent,  interest  be  allowed? 

In  this  question  A  must  be  the  amount  of  $10000  for  3  years 

7.  How  long  would  the  annuity  in  the  last  question  continue, 
on  condition  that  the  family  received  $1000  annually? 


MISCELLANEOUS    QUESTIONS 

1.  Four  men.  A,  B,  C  and  J,  bought  a  ship  for  $10428;  of 
which  B  paid  twice  as  much  as  A,  C  paid  as  much  as  A  and  B, 
and  D  paid  as  much  as  B  and  C.     How  much  did  each  pay  ? 

2.  A  person  bought  8  yards  of  cloth  for  «£3  2s,  giving  9s  a 
yard  for  a  part,  and  7s  a  yard  for  the  rest.  How  many  yards  did 
he  buy  at  each  price  ? 

3.  A  father  is  40  years  old,  and  his  son  8 ;  in  how  many 
years  will  the  father  be  three  times  as  old  as  the  son  ? 

4.  A  young  man  spends  ^  of  his  annual  income  for  board,  and 
^  as  much  for  clothes;  his  other  incidental  expenses  amount  to  J 
as  much  as  his  clothes,  and  yet  he  saves  $490  a  year.  What  is 
his  yearly  income? 

5.  A  person  had  spent  ^'of  his  life  in  England,  ^  of  it  on  the 
continent  of  Europe,  5  years  more  than  j^  of  it  in  Asia,  and  3 
years  more  than  ^  of  it  in  America.     How  old  was  he  ? 

6.  What  number  is  that,  from  which  if  5  be  subtracted,  and 
the  remainder  be  divided  by  2,  and  again  if  5  be  subtracted  from 
ihis  quotient,  and  the  remainder  be  divided  by  2,  it  will  leave  f*- 
of  the  number  itself? 


MISCELLANEOUS    Q  jESTIONS.  2S<I 

7  A  man  could  reap  a  field  of  wheat  in  5  days,  anc  his  son 
could  reap  it  in  20  days.  In  what  time  would  they  together  reap 
it? 

8.  If  a  certain  number  be  subtracted  from  100  and  120  re- 
spectively, ^  of  the  former  remainder  will  be  equal  to  ^  of  the 
"atter.     Required  the  number. 

9  There  is  a  rectangular  piece  of  land,  whose  length  ex- 
ceeds its  breadth  by  10  rods;  if  the  field  were  a  square  whose 
side  was  equal  to  its  present  length,  it  would  contain  400  square 
rods  more  than  it  now  contains.  Required  the  length  and 
breadth. 

10.  To  pay  a  debt  of  .£39  with  40  coins,  eagles  and  dollars, 
how  many  of  each  must  I  have,  the  dollar  being  6  shillings? 

11.  Two  men,  A  and  B,  had  together  $108;  the  former  spent 
^  and  the  latter  ^  of  what  he  had ;  and  the  amount  of  what  both 
spent  was  $32.     How  much  money  had  each  at  first  ? 

12.  A  merchant  commencing  business  with  a  certain  capital, 
lost  ^  of  it  the  first  year ;  but  the  next  year  he  gained  $700 ;  he 
thus  continued  alternately  losing  ^  of  what  he  had  at  the  time, 
and  gaining  $700,  until^  at  the  end  of  the  6th  year,  he  had  $350 
more  than  he  commenced  with  at  first.  With  what  capital  did 
he  commence? 

13.  A,  B  and  C  had  the  same  amount  of  mon^y  ;  A  gave 
away  $5,  and  spent  ^  of  the  remainder;  B  gave  away  $10,  and 
spent  -^  of  the  remainder;  C  gained  $10,  and  spent  y\y  of  what 
he  then  had;  after  which  they  had  together  $116.  How  much 
money  had  each  at  first  ? 

14.  A  father  leaves  to  his  three  sons  c£ 1 600,  in  the  following 
manner.  The  second  is  to  have  £200  less  than  the  eldest,  and 
£100  more  than  the  youngest.     Required  the  share  of  each. 

15.  Of  a  battalion  of  men,  |  of  the  whole  are  on  duty,  ^  are 
sick,  I  of  the  remainder  are  absent,  and  there  are  48  officers. 
How  many  persons  are  there  in  the  battalion  ? 

16.  A  and  B  found  a  purse  containing  dollars.     A  took  from 

25* 


»II4  MISCELLANEOUS    QUESTIONS. 

it  $2,  and  I  of  the  remainder ;  after  which  B  took  from  it  $3 
and  ^  of  the  remainder,  when  it  was  found  that  A  and  B  had 
taken  out  equal  sums.  How  much  money  was  there  in  the  purse 
at  first  ? 

17.  A  and  B  have  the  same  yearly  income ;  A  contracts  an 
annual  debt  amounting  to  j-  of  his ;  while  B  spends  only  f  of  his. 
At  the  end  of  10  years  B  lends  A  money  enough  to  pay  the  debt 
which  he  has  contracted  in  the  mean  time,  and  has  ^160  left. 
What  is  the  income  of  each  ? 

18.  A  gentleman  found,  that,  in  order  to  give  some  beggars. 
2s  6d  each,  he  would  want  3s ;  he  therefore  gave  2s  to  each,  and 
had  4s  left.  How  many  beggars  were  there,  and  how  much 
money  had  the  gentleman? 

19.  Find  a  number,  such,  that  whether  it  be  divided  into  two 
or  three  equal  parts,  the  continued  product  of  the  parts  shall  be 
of  the  same  value. 

20.  Divide  72  into  three  parts,  so  that  J  of  the  first  shall  be 
equal  to  the  second,  and  f  of  the  second  shall  be  equal  to  the 
.bird. 

21.  A  man  bought  6  bushels  of  wheat  and  3  bushels  of  rye  for 
$13;  he  afterwards  sold  4  bushels  of  wheat  and  7  bushels  of  rye 
at  the  same  rate  for  $13§.  How  many  shillings  were  given  a 
bushel  for  each? 

22.  There  is  a  certain  fraction,  to  the  numerator  which,  if  3 
be  added,  the  value  of  the  fraction  will  be  ^ ;  but  if  1  be  sub* 
tracted  from  the  denominator^  the  value  of  the  fraction  will  be  ^. 
What  is  the  fi-action  ? 

23.  There  is  a  number  consisting  of  two  digits.  The  sum  of 
the  digits  is  5 ;  and  if  9  be  added  to  the  number  itself,  the  digits 
will  be  inverted.     Required  the  number. 

24.  The  sum  of  two  numbers  is  37 ;  and  if  three  times  the  less 
be  subtracted  from  four  times  the  greater,  ^  of  the  difference 
ivill  be  6.     Required  the  numbers. 

25.  Separate  25  into  two  parts,  such  that  their  product  shall 


MISCELLANEOUS    QUESTIONS.  295 

26.  A  gambler  lost  ^  of  his  money,  and  then  won  3  shillings ; 
again  he  lost  J  of  what  he  then  had,  and  afterwards  won  2  shil- 
lings ;  lastly,  he  lost  ^  of  what  he  then  had,  and  found  that  he 
had  14  shillings  left.     How  much  money  had  he  at  first? 

27.  There  is  a  number  consisting  of  two  digits,  to  the  sum  of 
which,  if  7  be  added,  the  result  will  be  equal  to  three  times  the 
left  hand  digit;  but  if  18  be  subtracted  from  .ne  number  itself, 
the  digits  will  be  inverted.     Required  the  number. 

28.  Says  A  to  B,  give  me  $15  of  your  money,  and  I  shall  have 
as  much  as  ygu  will  have  left;  true,  says  B,  but  give  me  $10  of 
your  money,  and  I  shall  have  six  times  as  much  as  you  will  have 
left.     How  much  money  has  each? 

29.  A  vintner  has  two  casks  of  wine,  from  the  greater  of  which 
he  draws  15  gallons,  and  from  the  less  1 1  gallons,  and  the  quan- 
tities remaining  are  as  8  to  3.  After  the  casks  are  half  emptied, 
he  puts  10  gallons  of  water  into  each,  and  the  quantities  of 
liquor  then  in  them  are  as  9  to  5.  How  much  does  «ach  cask 
hold? 

30.  A  and  B  speculate  with  different  sums  of  money ;  A  gains 
£150,  and  B  loses  .£50 ;  then  A's  stock  is  to  B's  as  3  to  2.  But 
had  A  lost  £50,  and  B  gained  £100,  A's  stock  would  have 
been  to  B's  as  5  to  9.  Required  the  stock  with  which  each 
commenced. 

31.  If  a  certain  floor  were  5  feet  longer  and  4  feet  wider,  it 
would  contain  550  square  feet.  But  if  it  were  4  feet  longer  and 
5  feet  wider,  it  would  contain  192  square  feet  more  than  it  actu- 
ally does  contain.     Required  the  dimensions  of  the  floor. 

32.  If  A  work  3  days  and  B  4,  they  will  earn  $9 ;  if  A  work 
4  days  and  C  5,  they  will  earn  $14;  if  B  work  6  days  and  C  7, 
they  will  earn  $23.     Required  the  daily  wages  of  each. 

33.  Two  numbers  are  in  the  ratio  of  4  to  5,  and  the  difference 
of  their  second  powers  is  81.     What  are  these  numbers  ? 

34.  The  sum  of  two  numbers  is  18,  and  the  sum  of  their 
kquares  is  164.     Required  the  numbers. 

35.  What  two  numbers  are  those  whose  difference  is  7,  and 


896  MISCELLANEOUS    QUESTIONS. 

half  of  whose  product  increased  by  30,  is  equal  to  the  square  of 
the  less? 

36.  The  product  of  two  numbers  is  120.  Moreover,  if  2  bt 
added  to  the  less,  and  3  be  subtracted  from  the  greater,  the  pro- 
duct of  the  sum  and  difference  will  also  be  120.  Required  the 
numbers 

37.  A  certain  number  of  slieep  cost  ,£120;  if  8  sheep  more 
had  been  bought  for  the  same  sum,  each  would  have  cost  10s 
less.     Required  the  number  of  sTieep. 

38.  A,  B  and  C  had  together  MO ;  B,  C  and  D  had  £90  ;  C, 
D  and  A  had  .£80^  and  D,  A  and  B  had  <£70.  How  much 
money  had  each  ? 

39.  A  and  B  set  out  from  the  same  place,  and  at  the  same 
time,  to  travel  to  a  town  at  the  distance  of  300  miles.  A  goes  1 
mile  an  hour  more  than  B,  and  accomplishes  his  journey  10  hours 
sooner  than  B.     At  what  rate  does  each  travel  ? 

40.  A  number  consists  of  two  digits.  The  left  hand  digit  is 
three  times  the  right ;  and  if  12  be  subtracted  from  the  number, 
the  remainder  will  be  equal  to  the  square  of  the  left  hand  digit. 
Required  the  number. 

41.  A  starts  three  hours  and  20  minutes^  sooner  than  B,  and 
travels  uniformly  6  miles  an  hour.  B  starting  from  .the  same 
place  follows  at  the  rate  of  5  miles  the  first  hour,  6  miles  the  2d, 
7  miles  the  3d,  and  so  on.     In  what  time  will  B  overtake  A? 

42.  Two  men,  93  miles  apart,  set  out  at  the  same  time  to  meet. 
One  commences  at  3  miles  an  hour,  and  increases  his  rate  2 
miles  each  hour ;  the  other  commences  at  15  miles  an  hour,  and 
dijninishes  his  rate  3  miles  each  hour.  In  how  many  hours  will 
they  meet? 

43.  There  are  two  numbers,  such  that  27  times  the  greater  is 
equal  to  the  square  of  27  times  the  less ;  and  3  times  the  greater 
is  equaj  to  the  cube  of  3  times  the  less.  What  are  these  num- 
bers ? 

44.  A  and  B  each  bought  a  farm ;  A's  farm  exceeded  B's  by 
I  acres;  each  gave  as  m^ny  cents  per  acre  as  the^e  were  acres 


MISCELLANEOUS    QUESTIONS,  29? 

m  the  farm  unich  he  bought ;  and  both  together  paid  $816'16 
How  many  acres  did  each  buy  ? 

45  Find  two  numbers,  such  that  the  square  of  the  gi eater 
multiplied  by  the  less  shall  be  equal  to  100,  and  the  square  of 
the  less  multiplied  by  the  greater  shall  be  equal  to  80. 

46.  There  are  two  numbers,  whose  sum  is  to  the  greater  as 
40  is  to  the  less,  and  whose  sum  is  to  the  less  as  90  is  to  the 
greater.     Required  the  numbers. 

47.  A  rectangular  house  lot,  whose  length  exceeds  its  breadth 
by  50  feet,  contains  15000  square  feet.  Required  the  dimen- 
sions. . 

48.  The  sum  of  the  second  powers  of  two  numbers  is  244^ 
and  the  second  power  of  their  sum  is  484.  What  are  the  num- 
bers ? 

49.  The  breadth  of  a  rectangular  field  is  to  its  length  as  4  to 
5.  It  is  worth  twice  as  many  cents  per  square  rod  as  there  are 
rods  in  breadth,  and  the  worth  of  the  whole  is  $1600.  Required 
the  dimension? 

50.  The  sum  of  two  numbers  added  to  a  mean  proportional 
between  them  is  37 ;  and  the  sum  of  the  squares  of  the  numbers 
added  to  their  product  is  481.     Required  the  numbers. 

51.  The  sum  of  two  numbers  multiplied  by  their  product  is 
240 ;  and  their  difference  multiplied  by  their  product  is  48.  Re- 
quired the  numbers. 

52.  Separate  24  into  two  such  parts,  that  the  product  of  these 
parts  shall  be  to  the  sum  of  their  second  powers  as.  3  to  10. 

53.  The  sum  of  two  numbers  multiplied  by  the  square  of  their 
product  is  1800 ;  and  the  difference  of  the  numbers  multiplied 
by  the  square  of  their  product  is  450.    Required  the  numbers. 

54.  Find  two  numbers,  such  that  the  difference  of  their  squares 
shall  be  56;  and  ^  of  their  product  added  to  the  square  of  the 
less  shall  make  40. 

55.  In  a  certain  school,  the  number  studying  geometry  is  the 
square  root  of  the  whole  number  of  scholars;  f  of  the  whole 
learn  algebra ;  and  36  scholars  learn  arithntetic.     These  three 


4S98  MISCELLANEOUS    QUESTIONS. 

classes  constitute  the  whole  school.     Required  the  whole  num- 
ber of  scholars. 

56.  A  number,  consisting  of  two  digits,  being  multiplied  by 
the  left  hand  digit,  produces  46;  but  if  the  sum  of  the  digits  be 
multiplied  by  the  same  digit,  the  product  will  be  10.  What  is 
the  number? 

57.  There  are  two  rectangular  vats,  whose  cubical  coutenfa 
differ  by  20  feet.  The  bottom  of  each  is  a  square,  one  sides  of 
which  is  equal  to  the  depth  of  the  other  vat ;  and  the  capacities 
of  the  two  vats  are  as  4  to  5.     Required  the  depth  of  each. 

58.  What  number  is  that,  from  which  if  4  be  subtracted,  this 
remainder  shall  exceed  its  square  root  by  2? 

59.  What  number  is  that,  to  which  if  24  be  added,  and  the 
square  root  of  this  sum  be  extracted,  this  root  shall  be  less  than 
the  original  number  by  18? 

60.  A  board  fence  was  built  round  a  rectangular  court  to  a 
certain  height.  The  length  of  the  court  was  8  times  the  height 
of  the  fence  wanting  2  yards;  its  breadth,  6  times'  the  height  of 
the  fence  wanting  5  yards ;  and  the  area  of  the  court  exceeded 
that  of  the  fence  by  178  square  yards.  Required  the  height  of 
the  fence  and  the  dimensions  of  the  court. 

61.  A  man  bought  a  quantity  of  cloth  for  $60.  If  he  had 
bought  3  yards  more  for  the  sfime  money,  it  would  have  cost  $  I 
a  yard  less.     How  many  yards  did  he  buy  ? 

62.  Two  men,  A  and  B,  set  out  at  the  same  time,  the  former 
from  the  town  C,  and  the  latter  from  the  town  D,  and  travel  to 
wards  each  other.  When  they  met,  A  had  gone  30  miles  more 
than  B ;  and  according  to  the  rate  they  had  traveled,  A  could 
reach  D  in  4  days,  and  B  could  reach  C  in  9  days,  from  the  time 
of  meeting.     Required  the  distance  between  the  towns. 

63.  What  number  exceeds  its  square  root  by  20? 

64.  Two  retailers,  A  and  B,  jointly  invested  $500  in  business. 
A's  money  was  employed  5  months,  B's  only  2  months,  and  each 
received  $450  for  his  capital  and  gain.  How  much  money  did 
each  advance  f 


MISCELLANEOUS    QUESTIONS.  '299 

65.  Find  two  numbers,  whose  difference  added  to  the  differ 
ence  of  their  squares,  makes  150,  and  whose  sum  added  to  the 
sum  of  their  squares,  makes  330. 

66.  Find  a  number  consisting  of  three  digits,  such  that  the 
sum  of  the  squares  of  the  digits  shall  be  66 ;  that  the  square 
of  the  middle  digit  shall  exceed  the  product  of  the  other  two  by 
9 ;  and,  if  594  be  subtracted  from  the  number  itself,  the  digit! 
shall  be  inverted. 

67.  Five  gamesters,  A,  B,  C,  D,  and  E,  play  together,  on  coi> 
dition  that  he  who  loses,  shall  forfeit  to  all  the  rest  as  much 
money  as  they  already  have.  First  A  loses,  then  B,  then  C,  then 
D,  and  finally  E.  Yet,  at  the  end  of  the  fifth  game,  each  has 
left;  $32.     How  much  has  each  at  first? 

68.  A  and  B  sold  100  eggs,  and  each  received  the  same  sum. 
If  A  had  sold  as  many  as  B,  he  would  have  received  18  pence  for 
them;  and  if  B  had  sold  as  many  as  A,  he  would  have  received 
only  8  pence  for  them.     How  many  did  each  sell  ? 

69.  Separate  24  into  two  parts,  whose  product  shall  be  35 
times  their  difference. 

70.  What  two  numbers  are  those,  whose  product  is  4  times 
their  difference,  and  whose  product  multiplied  by  their  difference 
is  16? 

71.  The  sum  of  three  numbers  is  21 ;  if  the  first  be  subtracted 
fifom  the  second,  and  the  second  from  the  third,  the  latter  re- 
mainder will  exceed  the  former  by  3 ;  moreover,  the  sum  of  the 
squares  of  the  first  and  third  is  137.     Required  the  numbers. 

72.  There  are  two  rectangular  vessels,  which  together  hold 
180  cubic  feet ;  the  bottom  of  each  is  a  square  whose  side  is 
equal  to  the  height  of  the  other  vessel.  If  each  vessel  were  a 
cube  whose  side  was  equal  to  one  side  of  its  bottom,  the  two  ves- 
sels would  contain  189  cubic  feet.  Required  the  dimensions  of 
each. 

73.  There  are  two  numbers,  such  that  the  square  of  the 
greater,  multiplied  by  the  lesf?,  is  30  more  than  the  square  of  the 


STlO  MISCELLANEOUS    QUESTl  dN8. 

less,  multiplied  by  the  greater ;  moreover,  the  3d  power  of  the 
greater  exceeds  that  of  the  less  by  98.     What  are  the  numbers  1 

74.  What  number  is  that  whose  fourth  power  exceeds  ten 
limes  its  second  power  by  936  ? 

75.  Find  a  number,  such  that  if  its  square  root  be  increased 
by  4,  the  cube  root  of  the  sum  shall  be  2. 

76.  The  first  year  a  man  was  in  trade  he  doubled  his  money  , 
the  second  year  he  gained  $5  more  than  the  square  root  of  the 
number  of  dollars  he  had  at  the  commencement  of  that  year, 
when  he  received  a  legacy  of  as  many  dollars  as  were  equal  to 
the  square  of  the  number  he  then  had,  and  found  that  his  whole 
fortune  amounted  to  $  13340.  With  how  much  money  did  he 
commence  business? 

77.  If  the  sum  of  two  numbers  be  increased  by  2,  and  the  sec- 
ond power  of  this  result  be  added  to  the  sum  of  the  numbers,  the 
amount  will  be  154.  Moreover,  the  difference  between  the  sec- 
ond powers  of  the  two  numbers  is  40.     What  are  the  numbers  f 

78.  The  sum  of  three  numbers  in  progression  by  difference  is 
15  ;  and  the  sum  of  the  squares  of  the  extremes  is  58.  Required 
the  numbers.! 

79.  Four  numbers  are  in  progression  by  difference ;  the  sum 
of  the  squares  of  the  first  two  terms  is  10;  and  the  sum  of  the 
squares  of  the  last  two  terms  is  74.     What  are  the  numbers? 

80.  Find  three  numbers  in  progression  by  quotient,  whose  sum 
is  26,  and  the  sun\  of  whose  second  powers  is  364. 

81.  Four  numbers  are  in  progression  by  quotient;  the  sum  of 
the  first  two  is  30,  that  of  the  last  two  is  120.  Required  the 
numbers 

82.  Required  the  compound  irrterest  on  ^120,  for  10  years 
at  6  per  cent,  annually. 

83.  What  will  $300  amount  to  in  10  years,  at  compound  ii 
lerest  semi-annually,  the  yearly  rate  being  5  per  cent.  ? 


•gSlVBB.SIT7| 


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HFH    2<^      ^Vt{ 

LD21-100m-7,'33 

